Un article de Wikipédia, l'encyclopédie libre.
Les primitives des fonctions rationnelles se déduisent par celles de leur décomposition en éléments simples , donc des formules suivantes :
(On suppose a ≠ 0 .)
∫
(
a
x
+
b
)
n
d
x
=
1
(
n
+
1
)
a
(
a
x
+
b
)
n
+
1
+
C
{\displaystyle \int (ax+b)^{n}\,\mathrm {d} x={\frac {1}{(n+1)a}}(ax+b)^{n+1}+C}
pour tout entier relatif n différent de –1 (Formule de quadrature de Cavalieri (en) )
∫
1
a
x
+
b
d
x
=
1
a
ln
|
a
x
+
b
|
+
C
{\displaystyle \int {\frac {1}{ax+b}}\,\mathrm {d} x={\frac {1}{a}}\ln |ax+b|+C}
∫
1
a
x
2
+
b
x
+
c
d
x
=
{
2
−
(
b
2
−
4
a
c
)
arctan
2
a
x
+
b
−
(
b
2
−
4
a
c
)
+
C
si
b
2
−
4
a
c
<
0
−
2
2
a
x
+
b
+
C
si
b
2
−
4
a
c
=
0
1
b
2
−
4
a
c
ln
|
2
a
x
+
b
−
b
2
−
4
a
c
2
a
x
+
b
+
b
2
−
4
a
c
|
+
C
=
{
−
2
b
2
−
4
a
c
artanh
2
a
x
+
b
b
2
−
4
a
c
+
C
si
|
2
a
x
+
b
|
<
b
2
−
4
a
c
−
2
b
2
−
4
a
c
arcoth
2
a
x
+
b
b
2
−
4
a
c
+
C
sinon
si
b
2
−
4
a
c
>
0
{\displaystyle \int {\frac {1}{ax^{2}+bx+c}}\,\mathrm {d} x=\left\{{\begin{array}{lll}\displaystyle {\frac {2}{\sqrt {-(b^{2}-4ac)}}}\operatorname {arctan} {\frac {2ax+b}{\sqrt {-(b^{2}-4ac)}}}+C&{\text{ si }}&b^{2}-4ac<0\\[18pt]\displaystyle {\frac {-2}{2ax+b}}+C&{\text{ si }}&b^{2}-4ac=0\\[6pt]\displaystyle {\frac {1}{\sqrt {b^{2}-4ac}}}\ln \left|{\frac {2ax+b-{\sqrt {b^{2}-4ac}}}{2ax+b+{\sqrt {b^{2}-4ac}}}}\right|+C=\left\{{\begin{array}{ll}-{\frac {2}{\sqrt {b^{2}-4ac}}}\operatorname {artanh} {\frac {2ax+b}{\sqrt {b^{2}-4ac}}}+C&{\text{ si }}|2ax+b|<{\sqrt {b^{2}-4ac}}\\-{\frac {2}{\sqrt {b^{2}-4ac}}}\operatorname {arcoth} {\frac {2ax+b}{\sqrt {b^{2}-4ac}}}+C&{\text{ sinon }}\end{array}}\right.&{\text{ si }}&b^{2}-4ac>0\end{array}}\right.}
∫
x
a
x
2
+
b
x
+
c
d
x
=
1
2
a
ln
|
a
x
2
+
b
x
+
c
|
−
b
2
a
∫
1
a
x
2
+
b
x
+
c
d
x
{\displaystyle \int {\frac {x}{ax^{2}+bx+c}}\,\mathrm {d} x={\frac {1}{2a}}\ln |ax^{2}+bx+c|-{\frac {b}{2a}}\int {\frac {1}{ax^{2}+bx+c}}\,\mathrm {d} x}
Pour tout entier n ≥ 2 :
∫
1
(
a
x
2
+
b
x
+
c
)
n
d
x
=
−
2
a
x
+
b
(
n
−
1
)
(
b
2
−
4
a
c
)
(
a
x
2
+
b
x
+
c
)
n
−
1
−
2
(
2
n
−
3
)
a
(
n
−
1
)
(
b
2
−
4
a
c
)
∫
1
(
a
x
2
+
b
x
+
c
)
n
−
1
d
x
{\displaystyle {\begin{aligned}\int {\frac {1}{(ax^{2}+bx+c)^{n}}}\,\mathrm {d} x=-{\frac {2ax+b}{(n-1)(b^{2}-4ac)(ax^{2}+bx+c)^{n-1}}}-{\frac {2(2n-3)a}{(n-1)(b^{2}-4ac)}}\int {\frac {1}{(ax^{2}+bx+c)^{n-1}}}\,\mathrm {d} x\end{aligned}}}
∫
x
(
a
x
2
+
b
x
+
c
)
n
d
x
=
b
x
+
2
c
(
n
−
1
)
(
b
2
−
4
a
c
)
(
a
x
2
+
b
x
+
c
)
n
−
1
+
(
2
n
−
3
)
b
(
n
−
1
)
(
b
2
−
4
a
c
)
∫
1
(
a
x
2
+
b
x
+
c
)
n
−
1
d
x
{\displaystyle \int {\frac {x}{(ax^{2}+bx+c)^{n}}}\,\mathrm {d} x={\frac {bx+2c}{(n-1)(b^{2}-4ac)(ax^{2}+bx+c)^{n-1}}}+{\frac {(2n-3)b}{(n-1)(b^{2}-4ac)}}\int {\frac {1}{(ax^{2}+bx+c)^{n-1}}}\,\mathrm {d} x}
Table de primitives