Από τη Βικιπαίδεια, την ελεύθερη εγκυκλοπαίδεια
Ακολουθεί ένας κατάλογος ολοκληρωμάτων (αντιπαράγωγων ολοκληρωμάτων) ρητών συναρτήσεων [ 1] [ 2] [ 3] ρητή συνάρτηση μπορεί να αναλυθεί ως μερικό κλάσμα της συνάρτησης σε άθροισμα συναρτήσεων της μορφής :
a
(
x
−
b
)
n
{\displaystyle {\frac {a}{(x-b)^{n}}}}
a
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+
d
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n
.
{\displaystyle {\frac {ax+b}{\left((x-c)^{2}+d^{2}\right)^{n}}}.}
το οποίο μπορεί στη συνέχεια να ενσωματωθεί όρος προς όρο.
Για άλλους τύπους συναρτήσεων, βλ. λίστες ολοκληρωμάτων[ 4]
∫
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{\displaystyle \int {\frac {f'(x)}{f(x)}}\,dx=\ln \left|f(x)\right|+C}
∫
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d
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=
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a
arctan
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{\displaystyle \int {\frac {1}{x^{2}+a^{2}}}\,dx={\frac {1}{a}}\arctan {\frac {x}{a}}\,\!+C}
∫
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ln
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artanh
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ln
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(for
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arcoth
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ln
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(for
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{\displaystyle \int {\frac {1}{x^{2}-a^{2}}}\,dx={\frac {1}{2a}}\ln \left|{\frac {x-a}{x+a}}\right|+C={\begin{cases}\displaystyle -{\frac {1}{a}}\,\operatorname {artanh} {\frac {x}{a}}+C={\frac {1}{2a}}\ln {\frac {a-x}{a+x}}+C&{\text{(for }}|x|<|a|{\mbox{)}}\\[12pt]\displaystyle -{\frac {1}{a}}\,\operatorname {arcoth} {\frac {x}{a}}+C={\frac {1}{2a}}\ln {\frac {x-a}{x+a}}+C&{\text{(for }}|x|>|a|{\mbox{)}}\end{cases}}}
∫
1
a
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2
d
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1
2
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ln
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x
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−
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{
1
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artanh
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ln
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(for
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1
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arcoth
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ln
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(for
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{\displaystyle \int {\frac {1}{a^{2}-x^{2}}}\,dx={\frac {1}{2a}}\ln \left|{\frac {a+x}{a-x}}\right|+C={\begin{cases}\displaystyle {\frac {1}{a}}\,\operatorname {artanh} {\frac {x}{a}}+C={\frac {1}{2a}}\ln {\frac {a+x}{a-x}}+C&{\text{(for }}|x|<|a|{\mbox{)}}\\[12pt]\displaystyle {\frac {1}{a}}\,\operatorname {arcoth} {\frac {x}{a}}+C={\frac {1}{2a}}\ln {\frac {x+a}{x-a}}+C&{\text{(for }}|x|>|a|{\mbox{)}}\end{cases}}}
∫
d
x
x
2
n
+
1
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1
2
n
−
1
∑
k
=
1
2
n
−
1
sin
(
2
k
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1
2
n
π
)
arctan
[
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x
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cos
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π
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csc
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π
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−
1
2
cos
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π
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ln
|
x
2
−
2
x
cos
(
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k
−
1
2
n
π
)
+
1
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+
C
{\displaystyle \int {\frac {dx}{x^{2^{n}}+1}}={\frac {1}{2^{n-1}}}\sum _{k=1}^{2^{n-1}}\sin \left({\frac {2k-1}{2^{n}}}\pi \right)\arctan \left[\left(x-\cos \left({\frac {2k-1}{2^{n}}}\pi \right)\right)\csc \left({\frac {2k-1}{2^{n}}}\pi \right)\right]-{\frac {1}{2}}\cos \left({\frac {2k-1}{2^{n}}}\pi \right)\ln \left|x^{2}-2x\cos \left({\frac {2k-1}{2^{n}}}\pi \right)+1\right|+C}
Πολλές από τις ακόλουθες αντιπαραγωγές έχουν έναν όρο της μορφής ln |ax + b |. Επειδή αυτό είναι απροσδιόριστο όταν x = -b / a , η πιο γενική μορφή του αντιπαραγωγού αντικαθιστά τη σταθερά ολοκλήρωσης με μια τοπικά σταθερή συνάρτηση. [ 5]
∫
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ln
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<
0
1
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ln
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+
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>
0
{\displaystyle \int {\frac {1}{ax+b}}\,dx={\begin{cases}{\dfrac {1}{a}}\ln(-(ax+b))+C^{-}&ax+b<0\\{\dfrac {1}{a}}\ln(ax+b)+C^{+}&ax+b>0\end{cases}}}
συνήθως συντομογραφείται ως
∫
1
a
x
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d
x
=
1
a
ln
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a
x
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+
C
,
{\displaystyle \int {\frac {1}{ax+b}}\,dx={\frac {1}{a}}\ln \left|ax+b\right|+C,}
όπουC πρέπει να νοείται ως συμβολισμός για μια τοπικά σταθερή συνάρτηση του x . Αυτή η σύμβαση θα τηρηθεί στα επόμενα.
∫
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a
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d
x
=
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+
1
a
(
n
+
1
)
+
C
(για
n
≠
−
1
)
{\displaystyle \int (ax+b)^{n}\,dx={\frac {(ax+b)^{n+1}}{a(n+1)}}+C\qquad {\text{(για }}n\neq -1{\mbox{)}}}
∫
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b
d
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−
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a
2
ln
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{\displaystyle \int {\frac {x}{ax+b}}\,dx={\frac {x}{a}}-{\frac {b}{a^{2}}}\ln \left|ax+b\right|+C}
∫
m
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a
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m
a
2
ln
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+
C
{\displaystyle \int {\frac {mx+n}{ax+b}}\,dx={\frac {m}{a}}x+{\frac {an-bm}{a^{2}}}\ln \left|ax+b\right|+C}
∫
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ln
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{\displaystyle \int {\frac {x}{(ax+b)^{2}}}\,dx={\frac {b}{a^{2}(ax+b)}}+{\frac {1}{a^{2}}}\ln \left|ax+b\right|+C}
∫
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(for
n
∉
{
1
,
2
}
)
{\displaystyle \int {\frac {x}{(ax+b)^{n}}}\,dx={\frac {a(1-n)x-b}{a^{2}(n-1)(n-2)(ax+b)^{n-1}}}+C\qquad {\text{(for }}n\not \in \{1,2\}{\mbox{)}}}
∫
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(for
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∉
{
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}
)
{\displaystyle \int x(ax+b)^{n}\,dx={\frac {a(n+1)x-b}{a^{2}(n+1)(n+2)}}(ax+b)^{n+1}+C\qquad {\text{(for }}n\not \in \{-1,-2\}{\mbox{)}}}
∫
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ln
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+
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{\displaystyle \int {\frac {x^{2}}{ax+b}}\,dx={\frac {b^{2}\ln(\left|ax+b\right|)}{a^{3}}}+{\frac {ax^{2}-2bx}{2a^{2}}}+C}
∫
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ln
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{\displaystyle \int {\frac {x^{2}}{(ax+b)^{2}}}\,dx={\frac {1}{a^{3}}}\left(ax-2b\ln \left|ax+b\right|-{\frac {b^{2}}{ax+b}}\right)+C}
∫
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{\displaystyle \int {\frac {x^{2}}{(ax+b)^{3}}}\,dx={\frac {1}{a^{3}}}\left(\ln \left|ax+b\right|+{\frac {2b}{ax+b}}-{\frac {b^{2}}{2(ax+b)^{2}}}\right)+C}
∫
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(for
n
∉
{
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,
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,
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}
)
{\displaystyle \int {\frac {x^{2}}{(ax+b)^{n}}}\,dx={\frac {1}{a^{3}}}\left(-{\frac {(ax+b)^{3-n}}{(n-3)}}+{\frac {2b(ax+b)^{2-n}}{(n-2)}}-{\frac {b^{2}(ax+b)^{1-n}}{(n-1)}}\right)+C\qquad {\text{(for }}n\not \in \{1,2,3\}{\mbox{)}}}
∫
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ln
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{\displaystyle \int {\frac {1}{x(ax+b)}}\,dx=-{\frac {1}{b}}\ln \left|{\frac {ax+b}{x}}\right|+C}
∫
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{\displaystyle \int {\frac {1}{x^{2}(ax+b)}}\,dx=-{\frac {1}{bx}}+{\frac {a}{b^{2}}}\ln \left|{\frac {ax+b}{x}}\right|+C}
∫
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{\displaystyle \int {\frac {1}{x^{2}(ax+b)^{2}}}\,dx=-a\left({\frac {1}{b^{2}(ax+b)}}+{\frac {1}{ab^{2}x}}-{\frac {2}{b^{3}}}\ln \left|{\frac {ax+b}{x}}\right|\right)+C}
Για
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:
{\displaystyle a\neq 0:}
∫
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(for
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artanh
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(for
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arcoth
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(else)
(for
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(for
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)
{\displaystyle \int {\frac {1}{ax^{2}+bx+c}}dx={\begin{cases}\displaystyle {\frac {2}{\sqrt {4ac-b^{2}}}}\arctan {\frac {2ax+b}{\sqrt {4ac-b^{2}}}}+C&{\text{(for }}4ac-b^{2}>0{\mbox{)}}\\[12pt]\displaystyle {\frac {1}{\sqrt {b^{2}-4ac}}}\ln \left|{\frac {2ax+b-{\sqrt {b^{2}-4ac}}}{2ax+b+{\sqrt {b^{2}-4ac}}}}\right|+C={\begin{cases}\displaystyle -{\frac {2}{\sqrt {b^{2}-4ac}}}\,\operatorname {artanh} {\frac {2ax+b}{\sqrt {b^{2}-4ac}}}+C&{\text{(for }}|2ax+b|<{\sqrt {b^{2}-4ac}}{\mbox{)}}\\[6pt]\displaystyle -{\frac {2}{\sqrt {b^{2}-4ac}}}\,\operatorname {arcoth} {\frac {2ax+b}{\sqrt {b^{2}-4ac}}}+C&{\text{(else)}}\end{cases}}&{\text{(for }}4ac-b^{2}<0{\mbox{)}}\\[12pt]\displaystyle -{\frac {2}{2ax+b}}+C&{\text{(for }}4ac-b^{2}=0{\mbox{)}}\end{cases}}}
∫
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∫
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{\displaystyle \int {\frac {x}{ax^{2}+bx+c}}\,dx={\frac {1}{2a}}\ln \left|ax^{2}+bx+c\right|-{\frac {b}{2a}}\int {\frac {dx}{ax^{2}+bx+c}}+C}
∫
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arctan
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(for
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=
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artanh
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(for
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arcoth
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(else)
(for
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<
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)
m
2
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ln
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−
2
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n
−
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a
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+
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=
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a
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−
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n
−
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m
a
(
2
a
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+
b
)
+
C
(for
4
a
c
−
b
2
=
0
)
{\displaystyle \int {\frac {mx+n}{ax^{2}+bx+c}}\,dx={\begin{cases}\displaystyle {\frac {m}{2a}}\ln \left|ax^{2}+bx+c\right|+{\frac {2an-bm}{a{\sqrt {4ac-b^{2}}}}}\arctan {\frac {2ax+b}{\sqrt {4ac-b^{2}}}}+C&{\text{(for }}4ac-b^{2}>0{\mbox{)}}\\[12pt]\displaystyle {\frac {m}{2a}}\ln \left|ax^{2}+bx+c\right|+{\frac {2an-bm}{2a{\sqrt {b^{2}-4ac}}}}\ln \left|{\frac {2ax+b-{\sqrt {b^{2}-4ac}}}{2ax+b+{\sqrt {b^{2}-4ac}}}}\right|+C={\begin{cases}\displaystyle {\frac {m}{2a}}\ln \left|ax^{2}+bx+c\right|-{\frac {2an-bm}{a{\sqrt {b^{2}-4ac}}}}\,\operatorname {artanh} {\frac {2ax+b}{\sqrt {b^{2}-4ac}}}+C&{\text{(for }}|2ax+b|<{\sqrt {b^{2}-4ac}}{\mbox{)}}\\[6pt]\displaystyle {\frac {m}{2a}}\ln \left|ax^{2}+bx+c\right|-{\frac {2an-bm}{a{\sqrt {b^{2}-4ac}}}}\,\operatorname {arcoth} {\frac {2ax+b}{\sqrt {b^{2}-4ac}}}+C&{\text{(else)}}\end{cases}}&{\text{(for }}4ac-b^{2}<0{\mbox{)}}\\[12pt]\displaystyle {\frac {m}{2a}}\ln \left|ax^{2}+bx+c\right|-{\frac {2an-bm}{a(2ax+b)}}+C={\frac {m}{a}}\ln \left|x+{\frac {b}{2a}}\right|-{\frac {2an-bm}{a(2ax+b)}}+C&{\text{(for }}4ac-b^{2}=0{\mbox{)}}\end{cases}}}
∫
1
(
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x
2
+
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+
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)
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d
x
=
2
a
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+
b
(
n
−
1
)
(
4
a
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2
)
(
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+
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n
−
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+
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−
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2
a
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(
4
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−
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2
)
∫
1
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)
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−
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d
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+
C
{\displaystyle \int {\frac {1}{(ax^{2}+bx+c)^{n}}}\,dx={\frac {2ax+b}{(n-1)(4ac-b^{2})(ax^{2}+bx+c)^{n-1}}}+{\frac {(2n-3)2a}{(n-1)(4ac-b^{2})}}\int {\frac {1}{(ax^{2}+bx+c)^{n-1}}}\,dx+C}
∫
x
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(
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+
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−
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(
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2
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∫
1
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+
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+
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)
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−
1
d
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+
C
{\displaystyle \int {\frac {x}{(ax^{2}+bx+c)^{n}}}\,dx=-{\frac {bx+2c}{(n-1)(4ac-b^{2})(ax^{2}+bx+c)^{n-1}}}-{\frac {b(2n-3)}{(n-1)(4ac-b^{2})}}\int {\frac {1}{(ax^{2}+bx+c)^{n-1}}}\,dx+C}
∫
1
x
(
a
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2
+
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+
c
)
d
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=
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2
c
ln
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2
a
x
2
+
b
x
+
c
|
−
b
2
c
∫
1
a
x
2
+
b
x
+
c
d
x
+
C
{\displaystyle \int {\frac {1}{x(ax^{2}+bx+c)}}\,dx={\frac {1}{2c}}\ln \left|{\frac {x^{2}}{ax^{2}+bx+c}}\right|-{\frac {b}{2c}}\int {\frac {1}{ax^{2}+bx+c}}\,dx+C}
Τα προκύπτοντα ολοκληρώματα έχουν την ίδια μορφή με το αρχικό ολοκλήρωμα, οπότε αυτοί οι τύποι αναγωγής μπορούν να εφαρμοστούν επανειλημμένα για να οδηγήσουν τους εκθέτες m και p προς το 0.
Αυτοί οι τύποι αναγωγής μπορούν να χρησιμοποιηθούν για ολοκληρώματα που έχουν ακέραιους ή/και κλασματικούς εκθέτες.
∫
x
m
(
a
+
b
x
n
)
p
d
x
=
x
m
+
1
(
a
+
b
x
n
)
p
m
+
n
p
+
1
+
a
n
p
m
+
n
p
+
1
∫
x
m
(
a
+
b
x
n
)
p
−
1
d
x
{\displaystyle \int x^{m}\left(a+b\,x^{n}\right)^{p}dx={\frac {x^{m+1}\left(a+b\,x^{n}\right)^{p}}{m+n\,p+1}}\,+\,{\frac {a\,n\,p}{m+n\,p+1}}\int x^{m}\left(a+b\,x^{n}\right)^{p-1}dx}
∫
x
m
(
a
+
b
x
n
)
p
d
x
=
−
x
m
+
1
(
a
+
b
x
n
)
p
+
1
a
n
(
p
+
1
)
+
m
+
n
(
p
+
1
)
+
1
a
n
(
p
+
1
)
∫
x
m
(
a
+
b
x
n
)
p
+
1
d
x
{\displaystyle \int x^{m}\left(a+b\,x^{n}\right)^{p}dx=-{\frac {x^{m+1}\left(a+b\,x^{n}\right)^{p+1}}{a\,n(p+1)}}\,+\,{\frac {m+n(p+1)+1}{a\,n(p+1)}}\int x^{m}\left(a+b\,x^{n}\right)^{p+1}dx}
∫
x
m
(
a
+
b
x
n
)
p
d
x
=
x
m
+
1
(
a
+
b
x
n
)
p
m
+
1
−
b
n
p
m
+
1
∫
x
m
+
n
(
a
+
b
x
n
)
p
−
1
d
x
{\displaystyle \int x^{m}\left(a+b\,x^{n}\right)^{p}dx={\frac {x^{m+1}\left(a+b\,x^{n}\right)^{p}}{m+1}}\,-\,{\frac {b\,n\,p}{m+1}}\int x^{m+n}\left(a+b\,x^{n}\right)^{p-1}dx}
∫
x
m
(
a
+
b
x
n
)
p
d
x
=
x
m
−
n
+
1
(
a
+
b
x
n
)
p
+
1
b
n
(
p
+
1
)
−
m
−
n
+
1
b
n
(
p
+
1
)
∫
x
m
−
n
(
a
+
b
x
n
)
p
+
1
d
x
{\displaystyle \int x^{m}\left(a+b\,x^{n}\right)^{p}dx={\frac {x^{m-n+1}\left(a+b\,x^{n}\right)^{p+1}}{b\,n(p+1)}}\,-\,{\frac {m-n+1}{b\,n(p+1)}}\int x^{m-n}\left(a+b\,x^{n}\right)^{p+1}dx}
∫
x
m
(
a
+
b
x
n
)
p
d
x
=
x
m
−
n
+
1
(
a
+
b
x
n
)
p
+
1
b
(
m
+
n
p
+
1
)
−
a
(
m
−
n
+
1
)
b
(
m
+
n
p
+
1
)
∫
x
m
−
n
(
a
+
b
x
n
)
p
d
x
{\displaystyle \int x^{m}\left(a+b\,x^{n}\right)^{p}dx={\frac {x^{m-n+1}\left(a+b\,x^{n}\right)^{p+1}}{b(m+n\,p+1)}}\,-\,{\frac {a(m-n+1)}{b(m+n\,p+1)}}\int x^{m-n}\left(a+b\,x^{n}\right)^{p}dx}
∫
x
m
(
a
+
b
x
n
)
p
d
x
=
x
m
+
1
(
a
+
b
x
n
)
p
+
1
a
(
m
+
1
)
−
b
(
m
+
n
(
p
+
1
)
+
1
)
a
(
m
+
1
)
∫
x
m
+
n
(
a
+
b
x
n
)
p
d
x
{\displaystyle \int x^{m}\left(a+b\,x^{n}\right)^{p}dx={\frac {x^{m+1}\left(a+b\,x^{n}\right)^{p+1}}{a(m+1)}}\,-\,{\frac {b(m+n(p+1)+1)}{a(m+1)}}\int x^{m+n}\left(a+b\,x^{n}\right)^{p}dx}
Τα προκύπτοντα ολοκληρώματα έχουν την ίδια μορφή με το αρχικό ολοκλήρωμα, οπότε αυτοί οι τύποι αναγωγής μπορούν να εφαρμοστούν επανειλημμένα για να οδηγήσουν τους εκθέτες m , n και p προς το 0.
Αυτοί οι τύποι αναγωγής μπορούν να χρησιμοποιηθούν για ολοκληρώματα που έχουν ακέραιους ή/και κλασματικούς εκθέτες.
Ειδικές περιπτώσεις αυτών των τύπων αναγωγής μπορούν να χρησιμοποιηθούν για ολοκληρώματα της μορφής
(
a
+
b
x
)
m
(
c
+
d
x
)
n
(
e
+
f
x
)
p
{\displaystyle (a+b\,x)^{m}(c+d\,x)^{n}(e+f\,x)^{p}}
B στο 0.
∫
(
A
+
B
x
)
(
a
+
b
x
)
m
(
c
+
d
x
)
n
(
e
+
f
x
)
p
d
x
=
−
(
A
b
−
a
B
)
(
a
+
b
x
)
m
+
1
(
c
+
d
x
)
n
(
e
+
f
x
)
p
+
1
b
(
m
+
1
)
(
a
f
−
b
e
)
+
1
b
(
m
+
1
)
(
a
f
−
b
e
)
⋅
∫
(
b
c
(
m
+
1
)
(
A
f
−
B
e
)
+
(
A
b
−
a
B
)
(
n
d
e
+
c
f
(
p
+
1
)
)
+
d
(
b
(
m
+
1
)
(
A
f
−
B
e
)
+
f
(
n
+
p
+
1
)
(
A
b
−
a
B
)
)
x
)
(
a
+
b
x
)
m
+
1
(
c
+
d
x
)
n
−
1
(
e
+
f
x
)
p
d
x
{\displaystyle {\begin{aligned}&\int (A+B\,x)(a+b\,x)^{m}(c+d\,x)^{n}(e+f\,x)^{p}dx=-{\frac {(A\,b-a\,B)(a+b\,x)^{m+1}(c+d\,x)^{n}(e+f\,x)^{p+1}}{b(m+1)(a\,f-b\,e)}}\,+\,{\frac {1}{b(m+1)(a\,f-b\,e)}}\,\cdot \\&\qquad \int (b\,c(m+1)(A\,f-B\,e)+(A\,b-a\,B)(n\,d\,e+c\,f(p+1))+d(b(m+1)(A\,f-B\,e)+f(n+p+1)(A\,b-a\,B))x)(a+b\,x)^{m+1}(c+d\,x)^{n-1}(e+f\,x)^{p}dx\end{aligned}}}
∫
(
A
+
B
x
)
(
a
+
b
x
)
m
(
c
+
d
x
)
n
(
e
+
f
x
)
p
d
x
=
B
(
a
+
b
x
)
m
(
c
+
d
x
)
n
+
1
(
e
+
f
x
)
p
+
1
d
f
(
m
+
n
+
p
+
2
)
+
1
d
f
(
m
+
n
+
p
+
2
)
⋅
∫
(
A
a
d
f
(
m
+
n
+
p
+
2
)
−
B
(
b
c
e
m
+
a
(
d
e
(
n
+
1
)
+
c
f
(
p
+
1
)
)
)
+
(
A
b
d
f
(
m
+
n
+
p
+
2
)
+
B
(
a
d
f
m
−
b
(
d
e
(
m
+
n
+
1
)
+
c
f
(
m
+
p
+
1
)
)
)
)
x
)
(
a
+
b
x
)
m
−
1
(
c
+
d
x
)
n
(
e
+
f
x
)
p
d
x
{\displaystyle {\begin{aligned}&\int (A+B\,x)(a+b\,x)^{m}(c+d\,x)^{n}(e+f\,x)^{p}dx={\frac {B(a+b\,x)^{m}(c+d\,x)^{n+1}(e+f\,x)^{p+1}}{d\,f(m+n+p+2)}}\,+\,{\frac {1}{d\,f(m+n+p+2)}}\,\cdot \\&\qquad \int (A\,a\,d\,f(m+n+p+2)-B(b\,c\,e\,m+a(d\,e(n+1)+c\,f(p+1)))+(A\,b\,d\,f(m+n+p+2)+B(a\,d\,f\,m-b(d\,e(m+n+1)+c\,f(m+p+1))))x)(a+b\,x)^{m-1}(c+d\,x)^{n}(e+f\,x)^{p}dx\end{aligned}}}
∫
(
A
+
B
x
)
(
a
+
b
x
)
m
(
c
+
d
x
)
n
(
e
+
f
x
)
p
d
x
=
(
A
b
−
a
B
)
(
a
+
b
x
)
m
+
1
(
c
+
d
x
)
n
+
1
(
e
+
f
x
)
p
+
1
(
m
+
1
)
(
a
d
−
b
c
)
(
a
f
−
b
e
)
+
1
(
m
+
1
)
(
a
d
−
b
c
)
(
a
f
−
b
e
)
⋅
∫
(
(
m
+
1
)
(
A
(
a
d
f
−
b
(
c
f
+
d
e
)
)
+
B
b
c
e
)
−
(
A
b
−
a
B
)
(
d
e
(
n
+
1
)
+
c
f
(
p
+
1
)
)
−
d
f
(
m
+
n
+
p
+
3
)
(
A
b
−
a
B
)
x
)
(
a
+
b
x
)
m
+
1
(
c
+
d
x
)
n
(
e
+
f
x
)
p
d
x
{\displaystyle {\begin{aligned}&\int (A+B\,x)(a+b\,x)^{m}(c+d\,x)^{n}(e+f\,x)^{p}dx={\frac {(A\,b-a\,B)(a+b\,x)^{m+1}(c+d\,x)^{n+1}(e+f\,x)^{p+1}}{(m+1)(a\,d-b\,c)(a\,f-b\,e)}}\,+\,{\frac {1}{(m+1)(a\,d-b\,c)(a\,f-b\,e)}}\,\cdot \\&\qquad \int ((m+1)(A(a\,d\,f-b(c\,f+d\,e))+B\,b\,c\,e)-(A\,b-a\,B)(d\,e(n+1)+c\,f(p+1))-d\,f(m+n+p+3)(A\,b-a\,B)x)(a+b\,x)^{m+1}(c+d\,x)^{n}(e+f\,x)^{p}dx\end{aligned}}}
+ d x n q
Τα ολοκληρώματα που προκύπτουν έχουν την ίδια μορφή με το αρχικό ολοκλήρωμα, οπότε αυτοί οι τύποι αναγωγής μπορούν να εφαρμοστούν επανειλημμένα για να οδηγήσουν τους εκθέτες m , p και q προς το 0.
Οι εν λόγω τύποι αναγωγής μπορούν να χρησιμοποιηθούν για ολοκληρώματα που έχουν ακέραιους ή/και κλασματικούς εκθέτες.
Ειδικές περιπτώσεις αυτών των τύπων αναγωγής μπορούν να χρησιμοποιηθούν για ολοκληρώματα της μορφής
(
a
+
b
x
n
)
p
(
c
+
d
x
n
)
q
{\displaystyle \left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q}}
x
m
(
a
+
b
x
n
)
p
(
c
+
d
x
n
)
q
{\displaystyle x^{m}\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q}}
m ή/και το B στο 0.
∫
x
m
(
A
+
B
x
n
)
(
a
+
b
x
n
)
p
(
c
+
d
x
n
)
q
d
x
=
−
(
A
b
−
a
B
)
x
m
+
1
(
a
+
b
x
n
)
p
+
1
(
c
+
d
x
n
)
q
a
b
n
(
p
+
1
)
+
1
a
b
n
(
p
+
1
)
⋅
∫
x
m
(
c
(
A
b
n
(
p
+
1
)
+
(
A
b
−
a
B
)
(
m
+
1
)
)
+
d
(
A
b
n
(
p
+
1
)
+
(
A
b
−
a
B
)
(
m
+
n
q
+
1
)
)
x
n
)
(
a
+
b
x
n
)
p
+
1
(
c
+
d
x
n
)
q
−
1
d
x
{\displaystyle {\begin{aligned}&\int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q}dx=-{\frac {(A\,b-a\,B)x^{m+1}\left(a+b\,x^{n}\right)^{p+1}\left(c+d\,x^{n}\right)^{q}}{a\,b\,n(p+1)}}\,+\,{\frac {1}{a\,b\,n(p+1)}}\,\cdot \\&\qquad \int x^{m}\left(c(A\,b\,n(p+1)+(A\,b-a\,B)(m+1))+d(A\,b\,n(p+1)+(A\,b-a\,B)(m+n\,q+1))x^{n}\right)\left(a+b\,x^{n}\right)^{p+1}\left(c+d\,x^{n}\right)^{q-1}dx\end{aligned}}}
∫
x
m
(
A
+
B
x
n
)
(
a
+
b
x
n
)
p
(
c
+
d
x
n
)
q
d
x
=
B
x
m
+
1
(
a
+
b
x
n
)
p
+
1
(
c
+
d
x
n
)
q
b
(
m
+
n
(
p
+
q
+
1
)
+
1
)
+
1
b
(
m
+
n
(
p
+
q
+
1
)
+
1
)
⋅
∫
x
m
(
c
(
(
A
b
−
a
B
)
(
1
+
m
)
+
A
b
n
(
1
+
p
+
q
)
)
+
(
d
(
A
b
−
a
B
)
(
1
+
m
)
+
B
n
q
(
b
c
−
a
d
)
+
A
b
d
n
(
1
+
p
+
q
)
)
x
n
)
(
a
+
b
x
n
)
p
(
c
+
d
x
n
)
q
−
1
d
x
{\displaystyle {\begin{aligned}&\int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q}dx={\frac {B\,x^{m+1}\left(a+b\,x^{n}\right)^{p+1}\left(c+d\,x^{n}\right)^{q}}{b(m+n(p+q+1)+1)}}\,+\,{\frac {1}{b(m+n(p+q+1)+1)}}\,\cdot \\&\qquad \int x^{m}\left(c((A\,b-a\,B)(1+m)+A\,b\,n(1+p+q))+(d(A\,b-a\,B)(1+m)+B\,n\,q(b\,c-a\,d)+A\,b\,d\,n(1+p+q))\,x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q-1}dx\end{aligned}}}
∫
x
m
(
A
+
B
x
n
)
(
a
+
b
x
n
)
p
(
c
+
d
x
n
)
q
d
x
=
−
(
A
b
−
a
B
)
x
m
+
1
(
a
+
b
x
n
)
p
+
1
(
c
+
d
x
n
)
q
+
1
a
n
(
b
c
−
a
d
)
(
p
+
1
)
+
1
a
n
(
b
c
−
a
d
)
(
p
+
1
)
⋅
∫
x
m
(
c
(
A
b
−
a
B
)
(
m
+
1
)
+
A
n
(
b
c
−
a
d
)
(
p
+
1
)
+
d
(
A
b
−
a
B
)
(
m
+
n
(
p
+
q
+
2
)
+
1
)
x
n
)
(
a
+
b
x
n
)
p
+
1
(
c
+
d
x
n
)
q
d
x
{\displaystyle {\begin{aligned}&\int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q}dx=-{\frac {(A\,b-a\,B)x^{m+1}\left(a+b\,x^{n}\right)^{p+1}\left(c+d\,x^{n}\right)^{q+1}}{a\,n(b\,c-a\,d)(p+1)}}\,+\,{\frac {1}{a\,n(b\,c-a\,d)(p+1)}}\,\cdot \\&\qquad \int x^{m}\left(c(A\,b-a\,B)(m+1)+A\,n(b\,c-a\,d)(p+1)+d(A\,b-a\,B)(m+n(p+q+2)+1)x^{n}\right)\left(a+b\,x^{n}\right)^{p+1}\left(c+d\,x^{n}\right)^{q}dx\end{aligned}}}
∫
x
m
(
A
+
B
x
n
)
(
a
+
b
x
n
)
p
(
c
+
d
x
n
)
q
d
x
=
B
x
m
−
n
+
1
(
a
+
b
x
n
)
p
+
1
(
c
+
d
x
n
)
q
+
1
b
d
(
m
+
n
(
p
+
q
+
1
)
+
1
)
−
1
b
d
(
m
+
n
(
p
+
q
+
1
)
+
1
)
⋅
∫
x
m
−
n
(
a
B
c
(
m
−
n
+
1
)
+
(
a
B
d
(
m
+
n
q
+
1
)
−
b
(
−
B
c
(
m
+
n
p
+
1
)
+
A
d
(
m
+
n
(
p
+
q
+
1
)
+
1
)
)
)
x
n
)
(
a
+
b
x
n
)
p
(
c
+
d
x
n
)
q
d
x
{\displaystyle {\begin{aligned}&\int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q}dx={\frac {B\,x^{m-n+1}\left(a+b\,x^{n}\right)^{p+1}\left(c+d\,x^{n}\right)^{q+1}}{b\,d(m+n(p+q+1)+1)}}\,-\,{\frac {1}{b\,d(m+n(p+q+1)+1)}}\,\cdot \\&\qquad \int x^{m-n}\left(a\,B\,c(m-n+1)+(a\,B\,d(m+n\,q+1)-b(-B\,c(m+n\,p+1)+A\,d(m+n(p+q+1)+1)))x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q}dx\end{aligned}}}
∫
x
m
(
A
+
B
x
n
)
(
a
+
b
x
n
)
p
(
c
+
d
x
n
)
q
d
x
=
A
x
m
+
1
(
a
+
b
x
n
)
p
+
1
(
c
+
d
x
n
)
q
+
1
a
c
(
m
+
1
)
+
1
a
c
(
m
+
1
)
⋅
∫
x
m
+
n
(
a
B
c
(
m
+
1
)
−
A
(
b
c
+
a
d
)
(
m
+
n
+
1
)
−
A
n
(
b
c
p
+
a
d
q
)
−
A
b
d
(
m
+
n
(
p
+
q
+
2
)
+
1
)
x
n
)
(
a
+
b
x
n
)
p
(
c
+
d
x
n
)
q
d
x
{\displaystyle {\begin{aligned}&\int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q}dx={\frac {A\,x^{m+1}\left(a+b\,x^{n}\right)^{p+1}\left(c+d\,x^{n}\right)^{q+1}}{a\,c(m+1)}}\,+\,{\frac {1}{a\,c(m+1)}}\,\cdot \\&\qquad \int x^{m+n}\left(a\,B\,c(m+1)-A(b\,c+a\,d)(m+n+1)-A\,n(b\,c\,p+a\,d\,q)-A\,b\,d(m+n(p+q+2)+1)x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q}dx\end{aligned}}}
∫
x
m
(
A
+
B
x
n
)
(
a
+
b
x
n
)
p
(
c
+
d
x
n
)
q
d
x
=
A
x
m
+
1
(
a
+
b
x
n
)
p
+
1
(
c
+
d
x
n
)
q
a
(
m
+
1
)
−
1
a
(
m
+
1
)
⋅
∫
x
m
+
n
(
c
(
A
b
−
a
B
)
(
m
+
1
)
+
A
n
(
b
c
(
p
+
1
)
+
a
d
q
)
+
d
(
(
A
b
−
a
B
)
(
m
+
1
)
+
A
b
n
(
p
+
q
+
1
)
)
x
n
)
(
a
+
b
x
n
)
p
(
c
+
d
x
n
)
q
−
1
d
x
{\displaystyle {\begin{aligned}&\int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q}dx={\frac {A\,x^{m+1}\left(a+b\,x^{n}\right)^{p+1}\left(c+d\,x^{n}\right)^{q}}{a(m+1)}}\,-\,{\frac {1}{a(m+1)}}\,\cdot \\&\qquad \int x^{m+n}\left(c(A\,b-a\,B)(m+1)+A\,n(b\,c(p+1)+a\,d\,q)+d((A\,b-a\,B)(m+1)+A\,b\,n(p+q+1))x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q-1}dx\end{aligned}}}
∫
x
m
(
A
+
B
x
n
)
(
a
+
b
x
n
)
p
(
c
+
d
x
n
)
q
d
x
=
(
A
b
−
a
B
)
x
m
−
n
+
1
(
a
+
b
x
n
)
p
+
1
(
c
+
d
x
n
)
q
+
1
b
n
(
b
c
−
a
d
)
(
p
+
1
)
−
1
b
n
(
b
c
−
a
d
)
(
p
+
1
)
⋅
∫
x
m
−
n
(
c
(
A
b
−
a
B
)
(
m
−
n
+
1
)
+
(
d
(
A
b
−
a
B
)
(
m
+
n
q
+
1
)
−
b
n
(
B
c
−
A
d
)
(
p
+
1
)
)
x
n
)
(
a
+
b
x
n
)
p
+
1
(
c
+
d
x
n
)
q
d
x
{\displaystyle {\begin{aligned}&\int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q}dx={\frac {(A\,b-a\,B)x^{m-n+1}\left(a+b\,x^{n}\right)^{p+1}\left(c+d\,x^{n}\right)^{q+1}}{b\,n(b\,c-a\,d)(p+1)}}\,-\,{\frac {1}{b\,n(b\,c-a\,d)(p+1)}}\,\cdot \\&\qquad \int x^{m-n}\left(c(A\,b-a\,B)(m-n+1)+(d(A\,b-a\,B)(m+n\,q+1)-b\,n(B\,c-A\,d)(p+1))x^{n}\right)\left(a+b\,x^{n}\right)^{p+1}\left(c+d\,x^{n}\right)^{q}dx\end{aligned}}}
Τα προκύπτοντα ολοκληρώματα έχουν την ίδια μορφή με το αρχικό ολοκλήρωμα, οπότε αυτοί οι τύποι αναγωγής μπορούν να εφαρμοστούν επανειλημμένα για να οδηγήσουν τους εκθέτες m και p προς το 0.
Αυτοί οι τύποι αναγωγής μπορούν να χρησιμοποιηθούν για ολοκληρώματα που έχουν ακέραιους ή/και κλασματικούς εκθέτες.
Ειδικές περιπτώσεις αυτών των τύπων αναγωγής μπορούν να χρησιμοποιηθούν για ολοκληρώματα της μορφής
(
a
+
b
x
+
c
x
2
)
p
{\displaystyle \left(a+b\,x+c\,x^{2}\right)^{p}}
b
2
−
4
a
c
=
0
{\displaystyle b^{2}-4\,a\,c=0}
m to 0.
∫
(
d
+
e
x
)
m
(
a
+
b
x
+
c
x
2
)
p
d
x
=
(
d
+
e
x
)
m
+
1
(
a
+
b
x
+
c
x
2
)
p
e
(
m
+
1
)
−
p
(
d
+
e
x
)
m
+
2
(
b
+
2
c
x
)
(
a
+
b
x
+
c
x
2
)
p
−
1
e
2
(
m
+
1
)
(
m
+
2
p
+
1
)
+
p
(
2
p
−
1
)
(
2
c
d
−
b
e
)
e
2
(
m
+
1
)
(
m
+
2
p
+
1
)
∫
(
d
+
e
x
)
m
+
1
(
a
+
b
x
+
c
x
2
)
p
−
1
d
x
{\displaystyle \int (d+e\,x)^{m}\left(a+b\,x+c\,x^{2}\right)^{p}dx={\frac {(d+e\,x)^{m+1}\left(a+b\,x+c\,x^{2}\right)^{p}}{e(m+1)}}\,-\,{\frac {p(d+e\,x)^{m+2}(b+2c\,x)\left(a+b\,x+c\,x^{2}\right)^{p-1}}{e^{2}(m+1)(m+2p+1)}}\,+\,{\frac {p(2p-1)(2c\,d-b\,e)}{e^{2}(m+1)(m+2p+1)}}\int (d+e\,x)^{m+1}\left(a+b\,x+c\,x^{2}\right)^{p-1}dx}
∫
(
d
+
e
x
)
m
(
a
+
b
x
+
c
x
2
)
p
d
x
=
(
d
+
e
x
)
m
+
1
(
a
+
b
x
+
c
x
2
)
p
e
(
m
+
1
)
−
p
(
d
+
e
x
)
m
+
2
(
b
+
2
c
x
)
(
a
+
b
x
+
c
x
2
)
p
−
1
e
2
(
m
+
1
)
(
m
+
2
)
+
2
c
p
(
2
p
−
1
)
e
2
(
m
+
1
)
(
m
+
2
)
∫
(
d
+
e
x
)
m
+
2
(
a
+
b
x
+
c
x
2
)
p
−
1
d
x
{\displaystyle \int (d+e\,x)^{m}\left(a+b\,x+c\,x^{2}\right)^{p}dx={\frac {(d+e\,x)^{m+1}\left(a+b\,x+c\,x^{2}\right)^{p}}{e(m+1)}}\,-\,{\frac {p(d+e\,x)^{m+2}(b+2\,c\,x)\left(a+b\,x+c\,x^{2}\right)^{p-1}}{e^{2}(m+1)(m+2)}}\,+\,{\frac {2\,c\,p\,(2\,p-1)}{e^{2}(m+1)(m+2)}}\int (d+e\,x)^{m+2}\left(a+b\,x+c\,x^{2}\right)^{p-1}dx}
∫
(
d
+
e
x
)
m
(
a
+
b
x
+
c
x
2
)
p
d
x
=
−
e
(
m
+
2
p
+
2
)
(
d
+
e
x
)
m
(
a
+
b
x
+
c
x
2
)
p
+
1
(
p
+
1
)
(
2
p
+
1
)
(
2
c
d
−
b
e
)
+
(
d
+
e
x
)
m
+
1
(
b
+
2
c
x
)
(
a
+
b
x
+
c
x
2
)
p
(
2
p
+
1
)
(
2
c
d
−
b
e
)
+
e
2
m
(
m
+
2
p
+
2
)
(
p
+
1
)
(
2
p
+
1
)
(
2
c
d
−
b
e
)
∫
(
d
+
e
x
)
m
−
1
(
a
+
b
x
+
c
x
2
)
p
+
1
d
x
{\displaystyle \int (d+e\,x)^{m}\left(a+b\,x+c\,x^{2}\right)^{p}dx=-{\frac {e(m+2p+2)(d+e\,x)^{m}\left(a+b\,x+c\,x^{2}\right)^{p+1}}{(p+1)(2p+1)(2c\,d-b\,e)}}\,+\,{\frac {(d+e\,x)^{m+1}(b+2c\,x)\left(a+b\,x+c\,x^{2}\right)^{p}}{(2p+1)(2c\,d-b\,e)}}\,+\,{\frac {e^{2}m(m+2p+2)}{(p+1)(2p+1)(2c\,d-b\,e)}}\int (d+e\,x)^{m-1}\left(a+b\,x+c\,x^{2}\right)^{p+1}dx}
∫
(
d
+
e
x
)
m
(
a
+
b
x
+
c
x
2
)
p
d
x
=
−
e
m
(
d
+
e
x
)
m
−
1
(
a
+
b
x
+
c
x
2
)
p
+
1
2
c
(
p
+
1
)
(
2
p
+
1
)
+
(
d
+
e
x
)
m
(
b
+
2
c
x
)
(
a
+
b
x
+
c
x
2
)
p
2
c
(
2
p
+
1
)
+
e
2
m
(
m
−
1
)
2
c
(
p
+
1
)
(
2
p
+
1
)
∫
(
d
+
e
x
)
m
−
2
(
a
+
b
x
+
c
x
2
)
p
+
1
d
x
{\displaystyle \int (d+e\,x)^{m}\left(a+b\,x+c\,x^{2}\right)^{p}dx=-{\frac {e\,m(d+e\,x)^{m-1}\left(a+b\,x+c\,x^{2}\right)^{p+1}}{2c(p+1)(2p+1)}}\,+\,{\frac {(d+e\,x)^{m}(b+2c\,x)\left(a+b\,x+c\,x^{2}\right)^{p}}{2c(2p+1)}}\,+\,{\frac {e^{2}m(m-1)}{2c(p+1)(2p+1)}}\int (d+e\,x)^{m-2}\left(a+b\,x+c\,x^{2}\right)^{p+1}dx}
∫
(
d
+
e
x
)
m
(
a
+
b
x
+
c
x
2
)
p
d
x
=
(
d
+
e
x
)
m
+
1
(
a
+
b
x
+
c
x
2
)
p
e
(
m
+
2
p
+
1
)
−
p
(
2
c
d
−
b
e
)
(
d
+
e
x
)
m
+
1
(
b
+
2
c
x
)
(
a
+
b
x
+
c
x
2
)
p
−
1
2
c
e
2
(
m
+
2
p
)
(
m
+
2
p
+
1
)
+
p
(
2
p
−
1
)
(
2
c
d
−
b
e
)
2
2
c
e
2
(
m
+
2
p
)
(
m
+
2
p
+
1
)
∫
(
d
+
e
x
)
m
(
a
+
b
x
+
c
x
2
)
p
−
1
d
x
{\displaystyle \int (d+e\,x)^{m}\left(a+b\,x+c\,x^{2}\right)^{p}dx={\frac {(d+e\,x)^{m+1}\left(a+b\,x+c\,x^{2}\right)^{p}}{e(m+2p+1)}}\,-\,{\frac {p(2c\,d-b\,e)(d+e\,x)^{m+1}(b+2c\,x)\left(a+b\,x+c\,x^{2}\right)^{p-1}}{2c\,e^{2}(m+2p)(m+2p+1)}}\,+\,{\frac {p(2p-1)(2c\,d-b\,e)^{2}}{2c\,e^{2}(m+2p)(m+2p+1)}}\int (d+e\,x)^{m}\left(a+b\,x+c\,x^{2}\right)^{p-1}dx}
∫
(
d
+
e
x
)
m
(
a
+
b
x
+
c
x
2
)
p
d
x
=
−
2
c
e
(
m
+
2
p
+
2
)
(
d
+
e
x
)
m
+
1
(
a
+
b
x
+
c
x
2
)
p
+
1
(
p
+
1
)
(
2
p
+
1
)
(
2
c
d
−
b
e
)
2
+
(
d
+
e
x
)
m
+
1
(
b
+
2
c
x
)
(
a
+
b
x
+
c
x
2
)
p
(
2
p
+
1
)
(
2
c
d
−
b
e
)
+
2
c
e
2
(
m
+
2
p
+
2
)
(
m
+
2
p
+
3
)
(
p
+
1
)
(
2
p
+
1
)
(
2
c
d
−
b
e
)
2
∫
(
d
+
e
x
)
m
(
a
+
b
x
+
c
x
2
)
p
+
1
d
x
{\displaystyle \int (d+e\,x)^{m}\left(a+b\,x+c\,x^{2}\right)^{p}dx=-{\frac {2c\,e(m+2p+2)(d+e\,x)^{m+1}\left(a+b\,x+c\,x^{2}\right)^{p+1}}{(p+1)(2p+1)(2c\,d-b\,e)^{2}}}\,+\,{\frac {(d+e\,x)^{m+1}(b+2c\,x)\left(a+b\,x+c\,x^{2}\right)^{p}}{(2p+1)(2c\,d-b\,e)}}\,+\,{\frac {2c\,e^{2}(m+2p+2)(m+2p+3)}{(p+1)(2p+1)(2c\,d-b\,e)^{2}}}\int (d+e\,x)^{m}\left(a+b\,x+c\,x^{2}\right)^{p+1}dx}
∫
(
d
+
e
x
)
m
(
a
+
b
x
+
c
x
2
)
p
d
x
=
(
d
+
e
x
)
m
(
b
+
2
c
x
)
(
a
+
b
x
+
c
x
2
)
p
2
c
(
m
+
2
p
+
1
)
+
m
(
2
c
d
−
b
e
)
2
c
(
m
+
2
p
+
1
)
∫
(
d
+
e
x
)
m
−
1
(
a
+
b
x
+
c
x
2
)
p
d
x
{\displaystyle \int (d+e\,x)^{m}\left(a+b\,x+c\,x^{2}\right)^{p}dx={\frac {(d+e\,x)^{m}(b+2c\,x)\left(a+b\,x+c\,x^{2}\right)^{p}}{2c(m+2p+1)}}\,+\,{\frac {m(2c\,d-b\,e)}{2c(m+2p+1)}}\int (d+e\,x)^{m-1}\left(a+b\,x+c\,x^{2}\right)^{p}dx}
Τα προκύπτοντα ολοκληρώματα έχουν την ίδια μορφή με το αρχικό ολοκλήρωμα, οπότε αυτοί οι τύποι αναγωγής μπορούν να εφαρμοστούν επανειλημμένα για να οδηγήσουν τους εκθέτες m και p προς το 0.
Αυτοί οι τύποι αναγωγής μπορούν να χρησιμοποιηθούν για ολοκληρώματα που έχουν ακέραιους ή/και κλασματικούς εκθέτες.
Ειδικές περιπτώσεις αυτών των τύπων αναγωγής μπορούν να χρησιμοποιηθούν για ολοκληρώματα της μορφής
(
a
+
b
x
+
c
x
2
)
p
{\displaystyle \left(a+b\,x+c\,x^{2}\right)^{p}}
(
d
+
e
x
)
m
(
a
+
b
x
+
c
x
2
)
p
{\displaystyle (d+e\,x)^{m}\left(a+b\,x+c\,x^{2}\right)^{p}}
m και/ή B σε 0.
∫
(
d
+
e
x
)
m
(
A
+
B
x
)
(
a
+
b
x
+
c
x
2
)
p
d
x
=
(
d
+
e
x
)
m
+
1
(
A
e
(
m
+
2
p
+
2
)
−
B
d
(
2
p
+
1
)
+
e
B
(
m
+
1
)
x
)
(
a
+
b
x
+
c
x
2
)
p
e
2
(
m
+
1
)
(
m
+
2
p
+
2
)
+
1
e
2
(
m
+
1
)
(
m
+
2
p
+
2
)
p
⋅
∫
(
d
+
e
x
)
m
+
1
(
B
(
b
d
+
2
a
e
+
2
a
e
m
+
2
b
d
p
)
−
A
b
e
(
m
+
2
p
+
2
)
+
(
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{\displaystyle {\begin{aligned}&\int (d+e\,x)^{m}(A+B\,x)\left(a+b\,x+c\,x^{2}\right)^{p}dx={\frac {(d+e\,x)^{m+1}(A\,e(m+2p+2)-B\,d(2p+1)+e\,B(m+1)x)\left(a+b\,x+c\,x^{2}\right)^{p}}{e^{2}(m+1)(m+2p+2)}}\,+\,{\frac {1}{e^{2}(m+1)(m+2p+2)}}p\,\cdot \\&\qquad \int (d+e\,x)^{m+1}(B(b\,d+2a\,e+2a\,e\,m+2b\,d\,p)-A\,b\,e(m+2p+2)+(B(2c\,d+b\,e+b\,em+4c\,d\,p)-2A\,c\,e(m+2p+2))x)\left(a+b\,x+c\,x^{2}\right)^{p-1}dx\end{aligned}}}
∫
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{\displaystyle {\begin{aligned}&\int (d+e\,x)^{m}(A+B\,x)\left(a+b\,x+c\,x^{2}\right)^{p}dx={\frac {(d+e\,x)^{m}(A\,b-2a\,B-(b\,B-2A\,c)x)\left(a+b\,x+c\,x^{2}\right)^{p+1}}{(p+1)\left(b^{2}-4a\,c\right)}}\,+\,{\frac {1}{(p+1)\left(b^{2}-4a\,c\right)}}\,\cdot \\&\qquad \int (d+e\,x)^{m-1}(B(2a\,e\,m+b\,d(2p+3))-A(b\,e\,m+2c\,d(2p+3))+e(b\,B-2A\,c)(m+2p+3)x)\left(a+b\,x+c\,x^{2}\right)^{p+1}dx\end{aligned}}}
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{\displaystyle {\begin{aligned}&\int (d+e\,x)^{m}(A+B\,x)\left(a+b\,x+c\,x^{2}\right)^{p}dx={\frac {(d+e\,x)^{m+1}(A\,c\,e(m+2p+2)-B(c\,d+2c\,d\,p-b\,e\,p)+B\,c\,e(m+2p+1)x)\left(a+b\,x+c\,x^{2}\right)^{p}}{c\,e^{2}(m+2p+1)(m+2p+2)}}\,-\,{\frac {p}{c\,e^{2}(m+2p+1)(m+2p+2)}}\,\cdot \\&\qquad \int (d+e\,x)^{m}(A\,c\,e(b\,d-2a\,e)(m+2p+2)+B(a\,e(b\,e-2c\,d\,m+b\,e\,m)+b\,d(b\,e\,p-c\,d-2c\,d\,p))+\\&\qquad \qquad \left(A\,c\,e(2c\,d-b\,e)(m+2p+2)-B\left(-b^{2}e^{2}(m+p+1)+2c^{2}d^{2}(1+2p)+c\,e(b\,d(m-2p)+2a\,e(m+2p+1))\right)\right)x)\left(a+b\,x+c\,x^{2}\right)^{p-1}dx\end{aligned}}}
∫
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{\displaystyle {\begin{aligned}&\int (d+e\,x)^{m}(A+B\,x)\left(a+b\,x+c\,x^{2}\right)^{p}dx={\frac {(d+e\,x)^{m+1}\left(A\left(b\,c\,d-b^{2}e+2a\,c\,e\right)-a\,B(2c\,d-b\,e)+c(A(2c\,d-b\,e)-B(b\,d-2a\,e))x\right)\left(a+b\,x+c\,x^{2}\right)^{p+1}}{(p+1)\left(b^{2}-4a\,c\right)\left(c\,d^{2}-b\,d\,e+a\,e^{2}\right)}}\,+\\&\qquad {\frac {1}{(p+1)\left(b^{2}-4a\,c\right)\left(c\,d^{2}-b\,d\,e+a\,e^{2}\right)}}\,\cdot \\&\qquad \qquad \int (d+e\,x)^{m}(A\left(b\,c\,d\,e(2p-m+2)+b^{2}e^{2}(m+p+2)-2c^{2}d^{2}(3+2p)-2a\,c\,e^{2}(m+2p+3)\right)-\\&\qquad \qquad \qquad B(a\,e(b\,e-2c\,dm+b\,e\,m)+b\,d(-3c\,d+b\,e-2c\,d\,p+b\,e\,p))+c\,e(B(b\,d-2a\,e)-A(2c\,d-b\,e))(m+2p+4)x)\left(a+b\,x+c\,x^{2}\right)^{p+1}dx\end{aligned}}}
∫
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{\displaystyle {\begin{aligned}&\int (d+e\,x)^{m}(A+B\,x)\left(a+b\,x+c\,x^{2}\right)^{p}dx={\frac {B(d+e\,x)^{m}\left(a+b\,x+c\,x^{2}\right)^{p+1}}{c(m+2p+2)}}\,+\,{\frac {1}{c(m+2p+2)}}\,\cdot \\&\qquad \int (d+e\,x)^{m-1}(m(A\,c\,d-a\,B\,e)-d(b\,B-2A\,c)(p+1)+((B\,c\,d-b\,B\,e+A\,c\,e)m-e(b\,B-2A\,c)(p+1))x)\left(a+b\,x+c\,x^{2}\right)^{p}dx\end{aligned}}}
∫
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{\displaystyle {\begin{aligned}&\int (d+e\,x)^{m}(A+B\,x)\left(a+b\,x+c\,x^{2}\right)^{p}dx=-{\frac {(B\,d-A\,e)(d+e\,x)^{m+1}\left(a+b\,x+c\,x^{2}\right)^{p+1}}{(m+1)\left(c\,d^{2}-b\,d\,e+a\,e^{2}\right)}}\,+\,{\frac {1}{(m+1)\left(c\,d^{2}-b\,d\,e+a\,e^{2}\right)}}\,\cdot \\&\qquad \int (d+e\,x)^{m+1}((A\,c\,d-A\,b\,e+a\,B\,e)(m+1)+b(B\,d-A\,e)(p+1)+c(B\,d-A\,e)(m+2p+3)x)\left(a+b\,x+c\,x^{2}\right)^{p}dx\end{aligned}}}
∫
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{\displaystyle \int (d+e\,x)^{m}\left(a+b\,x+c\,x^{2}\right)^{p}dx=-{\frac {(d+e\,x)^{m+1}(b+2c\,x)\left(a+b\,x+c\,x^{2}\right)^{p}}{(m+1)(2c\,d-b\,e)}}\,+\,{\frac {2c(m+2p+2)}{(m+1)(2c\,d-b\,e)}}\int (d+e\,x)^{m+1}\left(a+b\,x+c\,x^{2}\right)^{p}dx}
Τα προκύπτοντα ολοκληρώματα έχουν την ίδια μορφή με το αρχικό ολοκλήρωμα, οπότε αυτοί οι τύποι αναγωγής μπορούν να εφαρμοστούν επανειλημμένα για να οδηγήσουν τους εκθέτες m και p προς το 0.
Αυτοί οι τύποι αναγωγής μπορούν να χρησιμοποιηθούν για ολοκληρώματα με ακέραιους ή/και κλασματικούς εκθέτες.
Ειδικές περιπτώσεις αυτών των τύπων αναγωγής μπορούν να χρησιμοποιηθούν για ολοκληρώματα της μορφής
(
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{\displaystyle \left(a+b\,x^{n}+c\,x^{2n}\right)^{p}}
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=
0
{\displaystyle b^{2}-4\,a\,c=0}
m to 0.
∫
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∫
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{\displaystyle \int x^{m}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx={\frac {x^{m+1}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}}{m+2n\,p+1}}\,+\,{\frac {n\,p\,x^{m+1}\left(2a+b\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p-1}}{(m+1)(m+2n\,p+1)}}\,-\,{\frac {b\,n^{2}p(2p-1)}{(m+1)(m+2n\,p+1)}}\int x^{m+n}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p-1}dx}
∫
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∫
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{\displaystyle \int x^{m}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx={\frac {(m+n(2p-1)+1)x^{m+1}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}}{(m+1)(m+n+1)}}\,+\,{\frac {n\,p\,x^{m+1}\left(2a+b\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p-1}}{(m+1)(m+n+1)}}\,+\,{\frac {2c\,p\,n^{2}(2p-1)}{(m+1)(m+n+1)}}\int x^{m+2n}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p-1}dx}
∫
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{\displaystyle \int x^{m}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx={\frac {(m+n(2p+1)+1)x^{m-n+1}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p+1}}{b\,n^{2}(p+1)(2p+1)}}\,-\,{\frac {x^{m+1}\left(b+2c\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}}{b\,n(2p+1)}}\,-\,{\frac {(m-n+1)(m+n(2p+1)+1)}{b\,n^{2}(p+1)(2p+1)}}\int x^{m-n}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p+1}dx}
∫
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{\displaystyle \int x^{m}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx=-{\frac {(m-3n-2n\,p+1)x^{m-2n+1}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p+1}}{2c\,n^{2}(p+1)(2p+1)}}\,-\,{\frac {x^{m-2n+1}\left(2a+b\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}}{2c\,n(2p+1)}}\,+\,{\frac {(m-n+1)(m-2n+1)}{2c\,n^{2}(p+1)(2p+1)}}\int x^{m-2n}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p+1}dx}
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{\displaystyle \int x^{m}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx={\frac {x^{m+1}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}}{m+2n\,p+1}}\,+\,{\frac {n\,p\,x^{m+1}\left(2a+b\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p-1}}{(m+2n\,p+1)(m+n(2p-1)+1)}}\,+\,{\frac {2a\,n^{2}p(2p-1)}{(m+2n\,p+1)(m+n(2p-1)+1)}}\int x^{m}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p-1}dx}
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{\displaystyle \int x^{m}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx=-{\frac {(m+n+2n\,p+1)x^{m+1}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p+1}}{2a\,n^{2}(p+1)(2p+1)}}\,-\,{\frac {x^{m+1}\left(2a+b\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}}{2a\,n(2p+1)}}\,+\,{\frac {(m+n(2p+1)+1)(m+2n(p+1)+1)}{2a\,n^{2}(p+1)(2p+1)}}\int x^{m}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p+1}dx}
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{\displaystyle \int x^{m}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx={\frac {x^{m-n+1}\left(b+2c\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}}{2c(m+2n\,p+1)}}\,-\,{\frac {b(m-n+1)}{2c(m+2n\,p+1)}}\int x^{m-n}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx}
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{\displaystyle \int x^{m}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx={\frac {x^{m+1}\left(b+2c\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}}{b(m+1)}}\,-\,{\frac {2c(m+n(2p+1)+1)}{b(m+1)}}\int x^{m+n}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx}
Τα προκύπτοντα ολοκληρώματα έχουν την ίδια μορφή με το αρχικό ολοκλήρωμα, οπότε αυτοί οι τύποι αναγωγής μπορούν να εφαρμοστούν επανειλημμένα για να οδηγήσουν τους εκθέτες m και p προς το 0.
Αυτοί οι τύποι αναγωγής μπορούν να χρησιμοποιηθούν για ολοκληρώματα που έχουν ακέραιους ή/και κλασματικούς εκθέτες.
Ειδικές περιπτώσεις αυτών των τύπων αναγωγής μπορούν να χρησιμοποιηθούν για ολοκληρώματα της μορφής
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{\displaystyle \left(a+b\,x^{n}+c\,x^{2n}\right)^{p}}
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{\displaystyle x^{m}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}}
m και/ή B σε 0.
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{\displaystyle {\begin{aligned}&\int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx={\frac {x^{m+1}\left(A(m+n(2p+1)+1)+B(m+1)x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}}{(m+1)(m+n(2p+1)+1)}}\,+\,{\frac {n\,p}{(m+1)(m+n(2p+1)+1)}}\,\cdot \\&\qquad \int x^{m+n}\left(2a\,B(m+1)-A\,b(m+n(2p+1)+1)+(b\,B(m+1)-2\,A\,c(m+n(2p+1)+1))x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p-1}dx\end{aligned}}}
∫
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{\displaystyle {\begin{aligned}&\int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx={\frac {x^{m-n+1}\left(A\,b-2a\,B-(b\,B-2A\,c)x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p+1}}{n(p+1)\left(b^{2}-4a\,c\right)}}\,+\,{\frac {1}{n(p+1)\left(b^{2}-4a\,c\right)}}\,\cdot \\&\qquad \int x^{m-n}\left((m-n+1)(2a\,B-A\,b)+(m+2n(p+1)+1)(b\,B-2A\,c)x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p+1}dx\end{aligned}}}
∫
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{\displaystyle {\begin{aligned}&\int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx={\frac {x^{m+1}\left(b\,B\,n\,p+A\,c(m+n(2p+1)+1)+B\,c(m+2n\,p+1)x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}}{c(m+2n\,p+1)(m+n(2p+1)+1)}}\,+\,{\frac {n\,p}{c(m+2n\,p+1)(m+n(2p+1)+1)}}\,\cdot \\&\qquad \int x^{m}\left(2a\,A\,c(m+n(2p+1)+1)-a\,b\,B(m+1)+\left(2a\,B\,c(m+2n\,p+1)+A\,b\,c(m+n(2p+1)+1)-b^{2}B(m+n\,p+1)\right)x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p-1}dx\end{aligned}}}
∫
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{\displaystyle {\begin{aligned}&\int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx=-{\frac {x^{m+1}\left(A\,b^{2}-a\,b\,B-2a\,A\,c+(A\,b-2a\,B)c\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p+1}}{a\,n(p+1)\left(b^{2}-4a\,c\right)}}\,+\,{\frac {1}{a\,n(p+1)\left(b^{2}-4a\,c\right)}}\,\cdot \\&\qquad \int x^{m}\left((m+n(p+1)+1)A\,b^{2}-a\,b\,B(m+1)-2(m+2n(p+1)+1)a\,A\,c+(m+n(2p+3)+1)(A\,b-2a\,B)c\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p+1}dx\end{aligned}}}
∫
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{\displaystyle {\begin{aligned}&\int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx={\frac {B\,x^{m-n+1}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p+1}}{c(m+n(2p+1)+1)}}\,-\,{\frac {1}{c(m+n(2p+1)+1)}}\,\cdot \\&\qquad \int x^{m-n}\left(a\,B(m-n+1)+(b\,B(m+n\,p+1)-A\,c(m+n(2p+1)+1))x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx\end{aligned}}}
∫
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{\displaystyle {\begin{aligned}&\int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx={\frac {A\,x^{m+1}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p+1}}{a(m+1)}}\,+\,{\frac {1}{a(m+1)}}\,\cdot \\&\qquad \int x^{m+n}\left(a\,B(m+1)-A\,b(m+n(p+1)+1)-A\,c(m+2n(p+1)+1)x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx\end{aligned}}}
Zwillinger, Daniel· Jeffrey, Alan (23 Φεβρουαρίου 2007). Table of Integrals, Series, and Products ISBN 978-0-08-047111-2 . Gradshteyn, I. S.· Ryzhik, I. M. (10 Μαΐου 2014). Table of Integrals, Series, and Products ISBN 978-1-4832-6564-3 . Lasserre, Jean-Bernard (2010). Moments, Positive Polynomials and Their Applications ISBN 978-1-84816-446-8 . Álvarez-Cónsul, Luis· Burgos-Gil, José Ignacio (24 Σεπτεμβρίου 2015). Feynman Amplitudes, Periods and Motives ISBN 978-1-4704-2247-9 . Singh, N. B. Higher Engineering Mathematics Blümlein, Johannes· Schneider, Carsten (26 Νοεμβρίου 2021). Anti-Differentiation and the Calculation of Feynman Amplitudes ISBN 978-3-030-80219-6 . Courant, Richard (23 Φεβρουαρίου 1988). Differential and Integral Calculus, Volume 1 ISBN 978-0-471-60842-4 . Gerdt, Vladimir P.· Koepf, Wolfram (10 Σεπτεμβρίου 2015). Computer Algebra in Scientific Computing: 17th International Workshop, CASC 2015, Aachen, Germany, September 14-18, 2015, Proceedings ISBN 978-3-319-24021-3 . Kerber, Manfred· Kohlhase, Michael (2 Απριλίου 2001). Symbolic Computation and Automated Reasoning: The CALCULEMUS-2000 Symposium ISBN 978-1-4398-6423-4 . Stewart, Seán M. (21 Δεκεμβρίου 2017). How to Integrate It: A Practical Guide to Finding Elementary Integrals ISBN 978-1-108-31414-5 .
Humanitarian Data Exchange(HDX) – The Humanitarian Data Exchange (HDX) is an open humanitarian data sharing platform managed by the United Nations Office for the Coordination of Humanitarian Affairs .NYC Open Data – free public data published by New York City agencies and other partners.Relational data set repository Αρχειοθετήθηκε 2018-03-07 στο Wayback Machine . Research Pipeline – a wiki/website with links to data sets on many different topicsStatLib–JASA Data Archive UCI – a machine learning repositoryUK Government Public Data World Bank Open Data – Free and open access to global development data by World Bank Apostol, Tom M. (29 Ιουνίου 2013). Introduction to Analytic Number Theory ISBN 978-1-4757-5579-4 . Miller, P. D. (2006), Applied Asymptotic Analysis American Mathematical Society , ISBN 9780821840788 , https://books.google.com/books?id=KQvqBwAAQBAJ Apostol, Thomas M. (1976), Introduction to Analytic Number Theory New York: Springer, ISBN 0-387-90163-9 , https://archive.org/details/introductiontoan00apos_0
![{\displaystyle \int {\frac {1}{x^{2}-a^{2}}}\,dx={\frac {1}{2a}}\ln \left|{\frac {x-a}{x+a}}\right|+C={\begin{cases}\displaystyle -{\frac {1}{a}}\,\operatorname {artanh} {\frac {x}{a}}+C={\frac {1}{2a}}\ln {\frac {a-x}{a+x}}+C&{\text{(for }}|x|<|a|{\mbox{)}}\\[12pt]\displaystyle -{\frac {1}{a}}\,\operatorname {arcoth} {\frac {x}{a}}+C={\frac {1}{2a}}\ln {\frac {x-a}{x+a}}+C&{\text{(for }}|x|>|a|{\mbox{)}}\end{cases}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f707d02dbc04ceb2d02ddb5bfd60ab31f45b6b55)
![{\displaystyle \int {\frac {1}{a^{2}-x^{2}}}\,dx={\frac {1}{2a}}\ln \left|{\frac {a+x}{a-x}}\right|+C={\begin{cases}\displaystyle {\frac {1}{a}}\,\operatorname {artanh} {\frac {x}{a}}+C={\frac {1}{2a}}\ln {\frac {a+x}{a-x}}+C&{\text{(for }}|x|<|a|{\mbox{)}}\\[12pt]\displaystyle {\frac {1}{a}}\,\operatorname {arcoth} {\frac {x}{a}}+C={\frac {1}{2a}}\ln {\frac {x+a}{x-a}}+C&{\text{(for }}|x|>|a|{\mbox{)}}\end{cases}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5d41f30fbfce8f00f5e1503b29b5e0b8415fadec)
![{\displaystyle \int {\frac {dx}{x^{2^{n}}+1}}={\frac {1}{2^{n-1}}}\sum _{k=1}^{2^{n-1}}\sin \left({\frac {2k-1}{2^{n}}}\pi \right)\arctan \left[\left(x-\cos \left({\frac {2k-1}{2^{n}}}\pi \right)\right)\csc \left({\frac {2k-1}{2^{n}}}\pi \right)\right]-{\frac {1}{2}}\cos \left({\frac {2k-1}{2^{n}}}\pi \right)\ln \left|x^{2}-2x\cos \left({\frac {2k-1}{2^{n}}}\pi \right)+1\right|+C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5086de865f6047c3f115bfbc3bd5ffde147a645c)
![{\displaystyle \int {\frac {1}{ax^{2}+bx+c}}dx={\begin{cases}\displaystyle {\frac {2}{\sqrt {4ac-b^{2}}}}\arctan {\frac {2ax+b}{\sqrt {4ac-b^{2}}}}+C&{\text{(for }}4ac-b^{2}>0{\mbox{)}}\\[12pt]\displaystyle {\frac {1}{\sqrt {b^{2}-4ac}}}\ln \left|{\frac {2ax+b-{\sqrt {b^{2}-4ac}}}{2ax+b+{\sqrt {b^{2}-4ac}}}}\right|+C={\begin{cases}\displaystyle -{\frac {2}{\sqrt {b^{2}-4ac}}}\,\operatorname {artanh} {\frac {2ax+b}{\sqrt {b^{2}-4ac}}}+C&{\text{(for }}|2ax+b|<{\sqrt {b^{2}-4ac}}{\mbox{)}}\\[6pt]\displaystyle -{\frac {2}{\sqrt {b^{2}-4ac}}}\,\operatorname {arcoth} {\frac {2ax+b}{\sqrt {b^{2}-4ac}}}+C&{\text{(else)}}\end{cases}}&{\text{(for }}4ac-b^{2}<0{\mbox{)}}\\[12pt]\displaystyle -{\frac {2}{2ax+b}}+C&{\text{(for }}4ac-b^{2}=0{\mbox{)}}\end{cases}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8c6e45e8f485cc92285459242e5edc389b0a4b3c)
![{\displaystyle \int {\frac {mx+n}{ax^{2}+bx+c}}\,dx={\begin{cases}\displaystyle {\frac {m}{2a}}\ln \left|ax^{2}+bx+c\right|+{\frac {2an-bm}{a{\sqrt {4ac-b^{2}}}}}\arctan {\frac {2ax+b}{\sqrt {4ac-b^{2}}}}+C&{\text{(for }}4ac-b^{2}>0{\mbox{)}}\\[12pt]\displaystyle {\frac {m}{2a}}\ln \left|ax^{2}+bx+c\right|+{\frac {2an-bm}{2a{\sqrt {b^{2}-4ac}}}}\ln \left|{\frac {2ax+b-{\sqrt {b^{2}-4ac}}}{2ax+b+{\sqrt {b^{2}-4ac}}}}\right|+C={\begin{cases}\displaystyle {\frac {m}{2a}}\ln \left|ax^{2}+bx+c\right|-{\frac {2an-bm}{a{\sqrt {b^{2}-4ac}}}}\,\operatorname {artanh} {\frac {2ax+b}{\sqrt {b^{2}-4ac}}}+C&{\text{(for }}|2ax+b|<{\sqrt {b^{2}-4ac}}{\mbox{)}}\\[6pt]\displaystyle {\frac {m}{2a}}\ln \left|ax^{2}+bx+c\right|-{\frac {2an-bm}{a{\sqrt {b^{2}-4ac}}}}\,\operatorname {arcoth} {\frac {2ax+b}{\sqrt {b^{2}-4ac}}}+C&{\text{(else)}}\end{cases}}&{\text{(for }}4ac-b^{2}<0{\mbox{)}}\\[12pt]\displaystyle {\frac {m}{2a}}\ln \left|ax^{2}+bx+c\right|-{\frac {2an-bm}{a(2ax+b)}}+C={\frac {m}{a}}\ln \left|x+{\frac {b}{2a}}\right|-{\frac {2an-bm}{a(2ax+b)}}+C&{\text{(for }}4ac-b^{2}=0{\mbox{)}}\end{cases}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/137aeb719faa0d412412ce2afb21f694747e79af)
