f
{\displaystyle f}
F
{\displaystyle F}
f
{\displaystyle f}
f
{\displaystyle f}
F
{\displaystyle F}
En raison de limitations techniques, la typographie souhaitable du titre, «
Exercice : Primitives 5Intégration en mathématiques/Exercices/Primitives 5 », n'a pu être restituée correctement ci-dessus.
f
(
x
)
=
ln
3
x
x
2
{\displaystyle f(x)={\frac {\ln ^{3}x}{x^{2}}}}
Solution
En utilisant que
∫
ln
k
x
x
2
d
x
=
[
−
ln
k
x
x
]
+
k
∫
ln
k
−
1
x
x
2
d
x
{\displaystyle \int {\frac {\ln ^{k}x}{x^{2}}}\,\mathrm {d} x=\left[{\frac {-\ln ^{k}x}{x}}\right]+k\int {\frac {\ln ^{k-1}x}{x^{2}}}\,\mathrm {d} x}
k
=
3
,
2
,
1
,
0
{\displaystyle k=3,2,1,0}
f
{\displaystyle f}
R
+
∗
{\displaystyle \mathbb {R} _{+}^{*}}
F
(
x
)
=
−
1
x
(
ln
3
x
+
3
ln
2
x
+
6
ln
x
+
6
)
{\displaystyle F(x)={\frac {-1}{x}}\left(\ln ^{3}x+3\ln ^{2}x+6\ln x+6\right)}
Plus généralement, une primitive sur
R
+
∗
{\displaystyle \mathbb {R} _{+}^{*}}
f
(
x
)
=
ln
n
x
x
2
{\displaystyle f(x)={\frac {\ln ^{n}x}{x^{2}}}}
F
(
x
)
=
−
1
x
∑
k
=
0
n
n
!
k
!
ln
k
x
{\displaystyle F(x)={\frac {-1}{x}}\sum _{k=0}^{n}{\frac {n!}{k!}}\ln ^{k}x}
f
(
x
)
=
(
x
3
+
x
)
e
−
3
x
{\displaystyle f(x)=\left(x^{3}+x\right)\mathrm {e} ^{-3x}}
Solution
∫
f
=
[
(
x
3
+
x
)
−
e
−
3
x
3
]
+
1
3
∫
(
3
x
2
+
1
)
e
−
3
x
d
x
=
[
(
x
3
+
x
+
3
x
2
+
1
3
)
−
e
−
3
x
3
]
+
2
3
∫
x
e
−
3
x
d
x
=
[
(
x
3
+
x
+
3
x
2
+
1
3
+
2
x
3
)
−
e
−
3
x
3
]
+
2
9
∫
e
−
3
x
d
x
=
[
(
x
3
+
x
+
3
x
2
+
1
3
+
2
x
3
+
2
9
)
−
e
−
3
x
3
]
=
[
(
x
3
+
x
2
+
5
x
3
+
5
9
)
−
e
−
3
x
3
]
{\displaystyle {\begin{aligned}\int f&=\left[\left(x^{3}+x\right){\frac {-\mathrm {e} ^{-3x}}{3}}\right]+{\frac {1}{3}}\int \left(3x^{2}+1\right)\mathrm {e} ^{-3x}\,\mathrm {d} x\\&=\left[\left(x^{3}+x+{\frac {3x^{2}+1}{3}}\right){\frac {-\mathrm {e} ^{-3x}}{3}}\right]+{\frac {2}{3}}\int x\mathrm {e} ^{-3x}\,\mathrm {d} x\\&=\left[\left(x^{3}+x+{\frac {3x^{2}+1}{3}}+{\frac {2x}{3}}\right){\frac {-\mathrm {e} ^{-3x}}{3}}\right]+{\frac {2}{9}}\int \mathrm {e} ^{-3x}\,\mathrm {d} x\\&=\left[\left(x^{3}+x+{\frac {3x^{2}+1}{3}}+{\frac {2x}{3}}+{\frac {2}{9}}\right){\frac {-\mathrm {e} ^{-3x}}{3}}\right]\\&=\left[\left(x^{3}+x^{2}+{\frac {5x}{3}}+{\frac {5}{9}}\right){\frac {-\mathrm {e} ^{-3x}}{3}}\right]\end{aligned}}}
donc une primitive de
f
{\displaystyle f}
R
{\displaystyle \mathbb {R} }
F
:
x
↦
(
x
3
+
x
2
+
5
x
3
+
5
9
)
−
e
−
3
x
3
{\displaystyle F:x\mapsto \left(x^{3}+x^{2}+{\frac {5x}{3}}+{\frac {5}{9}}\right){\frac {-\mathrm {e} ^{-3x}}{3}}}
f
(
x
)
=
(
x
2
−
x
+
6
)
e
−
5
x
{\displaystyle f(x)=\left(x^{2}-x+6\right)\mathrm {e} ^{-5x}}
Solution
∫
f
=
[
(
x
2
−
x
+
6
)
−
e
−
5
x
5
]
+
1
5
∫
(
2
x
−
1
)
e
−
5
x
d
x
=
[
(
x
2
−
x
+
6
+
2
x
−
1
5
)
−
e
−
5
x
5
]
+
2
25
∫
e
−
5
x
d
x
=
[
(
x
2
−
x
+
6
+
2
x
−
1
5
+
2
25
)
−
e
−
5
x
5
]
=
[
(
x
2
−
3
x
5
+
147
25
)
−
e
−
5
x
5
]
{\displaystyle {\begin{aligned}\int f&=\left[\left(x^{2}-x+6\right){\frac {-\mathrm {e} ^{-5x}}{5}}\right]+{\frac {1}{5}}\int \left(2x-1\right)\mathrm {e} ^{-5x}\,\mathrm {d} x\\&=\left[\left(x^{2}-x+6+{\frac {2x-1}{5}}\right){\frac {-\mathrm {e} ^{-5x}}{5}}\right]+{\frac {2}{25}}\int \mathrm {e} ^{-5x}\,\mathrm {d} x\\&=\left[\left(x^{2}-x+6+{\frac {2x-1}{5}}+{\frac {2}{25}}\right){\frac {-\mathrm {e} ^{-5x}}{5}}\right]\\&=\left[\left(x^{2}-{\frac {3x}{5}}+{\frac {147}{25}}\right){\frac {-\mathrm {e} ^{-5x}}{5}}\right]\end{aligned}}}
donc une primitive de
f
{\displaystyle f}
R
{\displaystyle \mathbb {R} }
F
:
x
↦
(
x
2
−
3
x
5
+
147
25
)
−
e
−
5
x
5
{\displaystyle F:x\mapsto \left(x^{2}-{\frac {3x}{5}}+{\frac {147}{25}}\right){\frac {-\mathrm {e} ^{-5x}}{5}}}
f
=
tan
+
tan
3
{\displaystyle f=\tan +\tan ^{3}}
f
(
x
)
=
1
x
4
e
1
x
3
{\displaystyle f(x)={\frac {1}{x^{4}}}\mathrm {e} ^{\frac {1}{x^{3}}}}
f
(
x
)
=
e
x
cos
(
2
x
)
{\displaystyle f(x)=\mathrm {e} ^{x}\cos(2x)}
g
(
x
)
=
e
x
sin
(
2
x
)
{\displaystyle g(x)=\mathrm {e} ^{x}\sin(2x)}
f
(
x
)
=
e
2
x
sin
x
{\displaystyle f(x)=\mathrm {e} ^{2x}\sin x}
Solution
∫
f
=
[
e
2
x
2
sin
2
x
]
−
∫
e
2
x
cos
2
x
d
x
=
[
e
2
x
2
(
sin
2
x
−
cos
2
x
)
]
−
∫
f
{\displaystyle {\begin{aligned}\int f&=\left[{\frac {\mathrm {e} ^{2x}}{2}}\sin 2x\right]-\int \mathrm {e} ^{2x}\cos 2x\,\mathrm {d} x\\&=\left[{\frac {\mathrm {e} ^{2x}}{2}}\left(\sin 2x-\cos 2x\right)\right]-\int f\end{aligned}}}
donc une primitive de
f
{\displaystyle f}
R
{\displaystyle \mathbb {R} }
F
:
x
↦
e
2
x
(
sin
2
x
−
cos
2
x
)
4
{\displaystyle F:x\mapsto {\frac {\mathrm {e} ^{2x}\left(\sin 2x-\cos 2x\right)}{4}}}
f
(
x
)
=
e
3
x
(
sin
2
x
−
cos
2
x
)
{\displaystyle f(x)=\mathrm {e} ^{3x}\left(\sin 2x-\cos 2x\right)}
Solution
∫
f
=
[
e
3
x
3
(
sin
2
x
−
cos
2
x
)
]
−
2
3
∫
e
3
x
(
cos
2
x
+
sin
2
x
)
d
x
=
[
e
3
x
3
(
sin
2
x
−
cos
2
x
−
2
3
(
cos
2
x
+
sin
2
x
)
)
]
+
4
9
∫
e
3
x
(
−
sin
2
x
+
cos
2
x
)
d
x
=
[
e
3
x
(
sin
2
x
−
5
cos
2
x
)
9
]
−
4
9
∫
f
{\displaystyle {\begin{aligned}\int f&=\left[{\frac {\mathrm {e} ^{3x}}{3}}\left(\sin 2x-\cos 2x\right)\right]-{\frac {2}{3}}\int \mathrm {e} ^{3x}\left(\cos 2x+\sin 2x\right)\,\mathrm {d} x\\&=\left[{\frac {\mathrm {e} ^{3x}}{3}}\left(\sin 2x-\cos 2x-{\frac {2}{3}}\left(\cos 2x+\sin 2x\right)\right)\right]+{\frac {4}{9}}\int \mathrm {e} ^{3x}\left(-\sin 2x+\cos 2x\right)\,\mathrm {d} x\\&=\left[{\frac {\mathrm {e} ^{3x}\left(\sin 2x-5\cos 2x\right)}{9}}\right]-{\frac {4}{9}}\int f\end{aligned}}}
donc une primitive de
f
{\displaystyle f}
R
{\displaystyle \mathbb {R} }
F
:
x
↦
e
3
x
(
sin
2
x
−
5
cos
2
x
)
13
{\displaystyle F:x\mapsto {\frac {\mathrm {e} ^{3x}\left(\sin 2x-5\cos 2x\right)}{13}}}
f
(
x
)
=
x
2
ln
(
1
+
x
)
{\displaystyle f(x)=x^{2}\ln(1+x)}
Solution
Sur
]
−
1
,
+
∞
[
{\displaystyle \left]-1,+\infty \right[}
∫
f
=
[
x
3
3
ln
(
x
+
1
)
]
−
1
3
∫
x
3
x
+
1
d
x
{\displaystyle \int f=\left[{\frac {x^{3}}{3}}\ln(x+1)\right]-{\frac {1}{3}}\int {\frac {x^{3}}{x+1}}\,\mathrm {d} x}
x
3
x
+
1
=
x
3
+
1
x
+
1
−
1
x
+
1
=
x
2
−
x
+
1
−
1
x
+
1
{\displaystyle {\frac {x^{3}}{x+1}}={\frac {x^{3}+1}{x+1}}-{\frac {1}{x+1}}=x^{2}-x+1-{\frac {1}{x+1}}}
donc une primitive de
f
{\displaystyle f}
]
−
1
,
+
∞
[
{\displaystyle \left]-1,+\infty \right[}
F
:
x
↦
1
3
(
(
x
3
+
1
)
ln
(
x
+
1
)
−
x
3
3
+
x
2
2
−
x
)
{\displaystyle F:x\mapsto {\frac {1}{3}}\left(\left(x^{3}+1\right)\ln(x+1)-{\frac {x^{3}}{3}}+{\frac {x^{2}}{2}}-x\right)}
f
(
x
)
=
cos
(
ln
x
)
{\displaystyle f(x)=\cos \left(\ln x\right)}
g
(
x
)
=
sin
(
ln
x
)
{\displaystyle g(x)=\sin \left(\ln x\right)}
Solution
∫
f
=
[
x
cos
(
ln
x
)
]
+
∫
g
{\displaystyle \int f=\left[x\cos \left(\ln x\right)\right]+\int g}
∫
g
=
[
x
sin
(
ln
x
)
]
−
∫
f
{\displaystyle \int g=\left[x\sin \left(\ln x\right)\right]-\int f}
donc (sur
]
0
,
+
∞
[
{\displaystyle \left]0,+\infty \right[}
f
{\displaystyle f}
F
:
x
↦
x
2
(
cos
(
ln
x
)
+
sin
(
ln
x
)
)
{\displaystyle F:x\mapsto {\frac {x}{2}}\left(\cos \left(\ln x\right)+\sin \left(\ln x\right)\right)}
et une primitive de
g
{\displaystyle g}
G
:
x
↦
x
2
(
sin
(
ln
x
)
−
cos
(
ln
x
)
)
{\displaystyle G:x\mapsto {\frac {x}{2}}\left(\sin \left(\ln x\right)-\cos \left(\ln x\right)\right)}
f
n
=
ln
n
{\displaystyle f_{n}=\ln ^{n}}
n
∈
N
{\displaystyle n\in \mathbb {N} }
Solution
Une intégration par parties donne comme primitive de
f
n
{\displaystyle f_{n}}
]
0
,
+
∞
[
{\displaystyle \left]0,+\infty \right[}
F
n
:
x
↦
x
f
n
(
x
)
−
n
F
n
−
1
(
x
)
{\displaystyle F_{n}:x\mapsto xf_{n}(x)-nF_{n-1}(x)}
F
n
(
x
)
=
x
∑
k
=
0
n
n
!
k
!
(
−
1
)
n
−
k
ln
k
(
x
)
{\displaystyle F_{n}(x)=x\sum _{k=0}^{n}{\frac {n!}{k!}}(-1)^{n-k}\ln ^{k}(x)}
Autre méthode :
F
k
(
x
)
=
G
k
(
ln
x
)
{\displaystyle F_{k}(x)=G_{k}(\ln x)}
ln
k
(
x
)
=
G
k
′
(
ln
x
)
/
x
{\displaystyle \ln ^{k}(x)=G_{k}'(\ln x)/x}
G
k
′
(
t
)
=
t
k
e
t
{\displaystyle G_{k}'(t)=t^{k}\operatorname {e} ^{t}}
G
k
(
t
)
=
e
t
H
k
(
t
)
{\displaystyle G_{k}(t)=e^{t}H_{k}(t)}
H
k
′
(
t
)
+
H
k
(
t
)
=
t
k
{\displaystyle H_{k}'(t)+H_{k}(t)=t^{k}}
H
k
(
t
)
=
a
k
t
k
+
…
+
a
0
{\displaystyle H_{k}(t)=a_{k}t^{k}+\ldots +a_{0}}
a
k
=
1
{\displaystyle a_{k}=1}
a
i
=
−
(
i
+
1
)
a
i
+
1
{\displaystyle a_{i}=-(i+1)a_{i+1}}
i
=
0
,
…
,
k
−
1
{\displaystyle i=0,\ldots ,k-1}
H
k
(
t
)
=
t
k
−
k
t
k
−
1
+
k
(
k
−
1
)
t
k
−
2
+
…
+
(
−
1
)
k
−
1
k
!
t
+
(
−
1
)
k
k
!
{\displaystyle H_{k}(t)=t^{k}-kt^{k-1}+k(k-1)t^{k-2}+\ldots +(-1)^{k-1}k!t+(-1)^{k}k!}
f
(
x
)
=
2
x
3
e
x
2
+
1
{\displaystyle f(x)=2x^{3}\operatorname {e} ^{x^{2}+1}}
f
(
x
)
=
ln
(
1
+
x
2
)
x
2
{\displaystyle f(x)={\frac {\ln(1+x^{2})}{x^{2}}}}
g
(
y
)
=
ln
(
1
+
1
/
y
2
)
{\displaystyle g(y)=\ln(1+1/y^{2})}
Déterminer les primitives des fonctions suivantes :
f
1
(
x
)
=
2
x
3
+
3
x
2
−
1
,
f
2
(
x
)
=
x
4
5
,
f
3
(
x
)
=
(
1
−
x
)
x
,
f
4
(
x
)
=
(
3
x
+
2
)
3
,
f
5
(
x
)
=
cos
(
4
x
+
1
)
,
f
6
(
x
)
=
x
1
−
x
2
{\displaystyle f_{1}(x)=2x^{3}+3x^{2}-1,\quad f_{2}(x)=x^{\frac {4}{5}},\quad f_{3}(x)=(1-x){\sqrt {x}},\quad f_{4}(x)=(3x+2)^{3},\quad f_{5}(x)=\cos(4x+1),\quad f_{6}(x)=x{\sqrt {1-x^{2}}}}
f
7
(
x
)
=
(
5
sin
x
+
2
)
3
cos
x
,
f
8
(
x
)
=
1
3
x
+
5
,
f
9
(
x
)
=
1
1
+
e
x
,
f
10
(
x
)
=
(
e
2
x
+
2
)
4
e
2
x
,
f
11
(
x
)
=
x
+
1
x
+
2
{\displaystyle f_{7}(x)=(5\sin x+2)^{3}\cos x,\quad f_{8}(x)={\frac {1}{3x+5}},\quad f_{9}(x)={\frac {1}{1+\mathrm {e} ^{x}}},\quad f_{10}(x)={(\mathrm {e} ^{2x}+2)^{4}\mathrm {e} ^{2x}},\quad f_{11}(x)={\frac {x+1}{x+2}}}
f
12
(
x
)
=
1
x
2
+
a
2
,
f
13
(
x
)
=
x
x
2
+
4
,
f
14
(
x
)
=
x
2
x
2
−
4
,
f
15
(
x
)
=
1
x
2
+
6
x
+
5
,
f
16
(
x
)
=
sin
2
x
,
f
17
(
x
)
=
cos
4
(
2
x
)
{\displaystyle f_{12}(x)={\frac {1}{x^{2}+a^{2}}},\quad f_{13}(x)={\frac {x}{x^{2}+4}},\quad f_{14}(x)={\frac {x^{2}}{x^{2}-4}},\quad f_{15}(x)={\frac {1}{x^{2}+6x+5}},\quad f_{16}(x)=\sin ^{2}x,\quad f_{17}(x)=\cos ^{4}(2x)}
f
18
(
x
)
=
cos
(
a
x
)
cos
(
b
x
)
et
g
18
(
x
)
=
cos
(
a
x
)
sin
(
b
x
)
avec
a
2
≠
b
2
,
f
19
=
arctan
,
f
20
(
x
)
=
ln
(
x
2
+
2
)
,
f
21
(
x
)
=
cos
x
,
f
22
(
x
)
=
ln
x
(
1
+
x
)
2
{\displaystyle f_{18}(x)=\cos(ax)\cos(bx){\text{ et }}g_{18}(x)=\cos(ax)\sin(bx){\text{ avec }}a^{2}\neq b^{2},\quad f_{19}=\arctan ,\quad f_{20}(x)=\ln(x^{2}+2),\quad f_{21}(x)=\cos {\sqrt {x}},\quad f_{22}(x)={\frac {\ln x}{(1+x)^{2}}}}
Solution
x
4
2
+
x
3
−
x
+
c
{\displaystyle {\frac {x^{4}}{2}}+x^{3}-x+c}
x
4
5
+
1
4
5
+
1
+
c
=
5
x
9
5
9
+
c
{\displaystyle {\frac {x^{{\frac {4}{5}}+1}}{{\frac {4}{5}}+1}}+c={\frac {5x^{\frac {9}{5}}}{9}}+c}
Changement de variable
x
=
y
2
{\displaystyle x=y^{2}}
∫
f
3
(
x
)
d
x
=
∫
(
1
−
y
2
)
y
2
y
d
y
=
2
y
3
3
−
2
y
5
5
+
c
=
2
x
3
2
3
−
2
x
5
2
5
+
c
=
2
x
(
x
3
−
x
2
5
)
+
c
{\displaystyle \int f_{3}(x)\;\mathrm {d} x=\int (1-y^{2})y\;2y\mathrm {d} y={\frac {2y^{3}}{3}}-{\frac {2y^{5}}{5}}+c={\frac {2x^{\frac {3}{2}}}{3}}-{\frac {2x^{\frac {5}{2}}}{5}}+c=2{\sqrt {x}}\left({\frac {x}{3}}-{\frac {x^{2}}{5}}\right)+c}
(
3
x
+
2
)
4
12
+
c
{\displaystyle {\frac {(3x+2)^{4}}{12}}+c}
sin
(
4
x
+
1
)
4
+
c
{\displaystyle {\frac {\sin(4x+1)}{4}}+c}
Changement de variable
y
=
1
−
x
2
{\displaystyle y={\sqrt {1-x^{2}}}}
∫
f
6
(
x
)
d
x
=
−
∫
y
2
d
y
=
−
y
3
3
+
c
=
−
(
1
−
x
2
)
3
2
3
+
c
{\displaystyle \int f_{6}(x)\;\mathrm {d} x=-\int y^{2}\mathrm {d} y=-{\frac {y^{3}}{3}}+c=-{\frac {(1-x^{2})^{\frac {3}{2}}}{3}}+c}
(
5
sin
x
+
2
)
4
20
+
c
{\displaystyle {\frac {(5\sin x+2)^{4}}{20}}+c}
ln
|
3
x
+
5
|
3
+
c
±
{\displaystyle {\frac {\ln |3x+5|}{3}}+c_{\pm }}
c
−
{\displaystyle c_{-}}
]
−
∞
,
−
5
3
[
{\displaystyle \left]-\infty ,-{\frac {5}{3}}\right[}
c
+
{\displaystyle c_{+}}
]
−
5
3
,
+
∞
[
{\displaystyle \left]-{\frac {5}{3}},+\infty \right[}
Changement de variable
y
=
e
x
{\displaystyle y=\mathrm {e} ^{x}}
∫
f
9
(
x
)
d
x
=
∫
d
y
y
(
1
+
y
)
=
∫
(
1
y
−
1
1
+
y
)
d
y
=
ln
|
y
|
−
ln
|
1
+
y
|
+
c
=
x
−
ln
(
1
+
e
x
)
+
c
{\displaystyle \int f_{9}(x)\;\mathrm {d} x=\int {\frac {\mathrm {d} y}{y(1+y)}}=\int \left({\frac {1}{y}}-{\frac {1}{1+y}}\right)\;\mathrm {d} y=\ln |y|-\ln |1+y|+c=x-\ln(1+\mathrm {e} ^{x})+c}
(
e
2
x
+
2
)
5
10
+
c
{\displaystyle {\frac {(\mathrm {e} ^{2x}+2)^{5}}{10}}+c}
∫
(
1
−
1
x
+
2
)
d
x
=
x
−
ln
|
x
+
2
|
+
c
±
{\displaystyle \int \left(1-{\frac {1}{x+2}}\right)\;\mathrm {d} x=x-\ln |x+2|+c_{\pm }}
arctan
x
a
a
+
c
{\displaystyle {\frac {\arctan {\frac {x}{a}}}{a}}+c}
ln
(
x
2
+
4
)
2
+
c
{\displaystyle {\frac {\ln(x^{2}+4)}{2}}+c}
f
14
(
x
)
=
1
+
1
x
−
2
−
1
x
+
2
{\displaystyle f_{14}(x)=1+{\frac {1}{x-2}}-{\frac {1}{x+2}}}
F
14
(
x
)
=
x
+
ln
|
x
−
2
|
−
ln
|
x
+
2
|
+
c
j
{\displaystyle F_{14}(x)=x+\ln |x-2|-\ln |x+2|+c_{j}}
f
15
(
x
)
=
1
4
(
1
x
+
1
−
1
x
+
5
)
{\displaystyle f_{15}(x)={\frac {1}{4}}\left({\frac {1}{x+1}}-{\frac {1}{x+5}}\right)}
F
15
(
x
)
=
1
4
(
ln
|
x
+
1
|
−
ln
|
x
+
5
|
)
+
c
j
{\displaystyle F_{15}(x)={\frac {1}{4}}\left(\ln |x+1|-\ln |x+5|\right)+c_{j}}
f
16
(
x
)
=
1
−
cos
(
2
x
)
2
{\displaystyle f_{16}(x)={\frac {1-\cos(2x)}{2}}}
F
16
(
x
)
=
x
2
−
sin
(
2
x
)
4
+
c
{\displaystyle F_{16}(x)={\frac {x}{2}}-{\frac {\sin(2x)}{4}}+c}
f
17
(
x
)
=
1
8
cos
(
8
x
)
+
1
2
cos
(
4
x
)
+
3
8
{\displaystyle f_{17}(x)={\frac {1}{8}}\cos(8x)+{\frac {1}{2}}\cos(4x)+{\frac {3}{8}}}
F
17
(
x
)
=
1
64
sin
(
4
x
)
+
1
8
sin
(
2
x
)
+
3
x
8
+
c
{\displaystyle F_{17}(x)={\frac {1}{64}}\sin(4x)+{\frac {1}{8}}\sin(2x)+{\frac {3x}{8}}+c}
f
18
(
x
)
=
1
2
cos
(
(
a
+
b
)
x
)
+
1
2
cos
(
(
a
−
b
)
x
)
{\displaystyle f_{18}(x)={\frac {1}{2}}\cos {\big (}(a+b)x{\big )}+{\frac {1}{2}}\cos {\big (}(a-b)x{\big )}}
F
18
(
x
)
=
1
2
(
a
+
b
)
sin
(
(
a
+
b
)
x
)
+
1
2
(
a
−
b
)
sin
(
(
a
−
b
)
x
)
+
c
{\displaystyle F_{18}(x)={\frac {1}{2(a+b)}}\sin {\big (}(a+b)x{\big )}+{\frac {1}{2(a-b)}}\sin {\big (}(a-b)x{\big )}+c}
g
18
(
x
)
=
1
2
sin
(
(
a
+
b
)
x
)
−
1
2
sin
(
(
a
−
b
)
x
)
{\displaystyle g_{18}(x)={\frac {1}{2}}\sin {\big (}(a+b)x{\big )}-{\frac {1}{2}}\sin {\big (}(a-b)x{\big )}}
G
18
(
x
)
=
1
2
(
a
−
b
)
cos
(
(
a
−
b
)
x
)
−
1
2
(
a
+
b
)
cos
(
(
a
+
b
)
x
)
+
c
{\displaystyle G_{18}(x)={\frac {1}{2(a-b)}}\cos {\big (}(a-b)x{\big )}-{\frac {1}{2(a+b)}}\cos {\big (}(a+b)x{\big )}+c}
[
x
arctan
x
]
−
∫
x
1
+
x
2
d
x
=
x
arctan
x
−
ln
(
1
+
x
2
)
2
+
c
{\displaystyle \left[x\arctan x\right]-\int {\frac {x}{1+x^{2}}}\;\mathrm {d} x=x\arctan x-{\frac {\ln(1+x^{2})}{2}}+c}
[
x
ln
(
x
2
+
2
)
]
−
2
∫
x
2
x
2
+
2
d
x
=
[
x
ln
(
x
2
+
2
)
]
−
2
∫
(
1
−
2
x
2
+
2
)
d
x
=
x
ln
(
x
2
+
2
)
−
2
x
+
2
2
arctan
x
2
+
c
{\displaystyle \left[x\ln(x^{2}+2)\right]-2\int {\frac {x^{2}}{x^{2}+2}}\;\mathrm {d} x=\left[x\ln(x^{2}+2)\right]-2\int \left(1-{\frac {2}{x^{2}+2}}\right)\;\mathrm {d} x=x\ln(x^{2}+2)-2x+2{\sqrt {2}}\arctan {\frac {x}{\sqrt {2}}}+c}
∫
cos
x
d
x
=
∫
cos
s
2
s
d
s
{\displaystyle \int \cos {\sqrt {x}}\;\mathrm {d} x=\int \cos s\;2s\mathrm {d} s}
s
cos
s
{\displaystyle s\cos s}
Intégration en mathématiques/Exercices/Primitives 3#Exercice 6-2 ) est
s
sin
s
+
cos
s
{\displaystyle s\sin s+\cos s}
F
21
(
x
)
=
2
(
x
sin
x
+
cos
x
)
+
c
{\displaystyle F_{21}(x)=2({\sqrt {x}}\sin {\sqrt {x}}+\cos {\sqrt {x}})+c}
[
−
ln
x
1
+
x
]
+
∫
1
x
(
1
+
x
)
d
x
=
[
−
ln
x
1
+
x
]
+
∫
(
1
x
−
1
1
+
x
)
d
x
{\displaystyle \left[{\frac {-\ln x}{1+x}}\right]+\int {\frac {1}{x(1+x)}}\;\mathrm {d} x=\left[{\frac {-\ln x}{1+x}}\right]+\int \left({\frac {1}{x}}-{\frac {1}{1+x}}\right)\;\mathrm {d} x}
F
22
(
x
)
=
−
ln
x
1
+
x
+
ln
x
1
+
x
+
c
=
x
ln
x
1
+
x
−
ln
(
1
+
x
)
+
c
{\displaystyle F_{22}(x)={\frac {-\ln x}{1+x}}+\ln {\frac {x}{1+x}}+c={\frac {x\ln x}{1+x}}-\ln(1+x)+c}
![{\displaystyle \int {\frac {\ln ^{k}x}{x^{2}}}\,\mathrm {d} x=\left[{\frac {-\ln ^{k}x}{x}}\right]+k\int {\frac {\ln ^{k-1}x}{x^{2}}}\,\mathrm {d} x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/51f2c6db0f9bd2ce5f25cb803df94aa0c11c1195)
![{\displaystyle {\begin{aligned}\int f&=\left[\left(x^{3}+x\right){\frac {-\mathrm {e} ^{-3x}}{3}}\right]+{\frac {1}{3}}\int \left(3x^{2}+1\right)\mathrm {e} ^{-3x}\,\mathrm {d} x\\&=\left[\left(x^{3}+x+{\frac {3x^{2}+1}{3}}\right){\frac {-\mathrm {e} ^{-3x}}{3}}\right]+{\frac {2}{3}}\int x\mathrm {e} ^{-3x}\,\mathrm {d} x\\&=\left[\left(x^{3}+x+{\frac {3x^{2}+1}{3}}+{\frac {2x}{3}}\right){\frac {-\mathrm {e} ^{-3x}}{3}}\right]+{\frac {2}{9}}\int \mathrm {e} ^{-3x}\,\mathrm {d} x\\&=\left[\left(x^{3}+x+{\frac {3x^{2}+1}{3}}+{\frac {2x}{3}}+{\frac {2}{9}}\right){\frac {-\mathrm {e} ^{-3x}}{3}}\right]\\&=\left[\left(x^{3}+x^{2}+{\frac {5x}{3}}+{\frac {5}{9}}\right){\frac {-\mathrm {e} ^{-3x}}{3}}\right]\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3d2fa74dbb3622ff8b4f73ae2327b659a89e36d0)
![{\displaystyle {\begin{aligned}\int f&=\left[\left(x^{2}-x+6\right){\frac {-\mathrm {e} ^{-5x}}{5}}\right]+{\frac {1}{5}}\int \left(2x-1\right)\mathrm {e} ^{-5x}\,\mathrm {d} x\\&=\left[\left(x^{2}-x+6+{\frac {2x-1}{5}}\right){\frac {-\mathrm {e} ^{-5x}}{5}}\right]+{\frac {2}{25}}\int \mathrm {e} ^{-5x}\,\mathrm {d} x\\&=\left[\left(x^{2}-x+6+{\frac {2x-1}{5}}+{\frac {2}{25}}\right){\frac {-\mathrm {e} ^{-5x}}{5}}\right]\\&=\left[\left(x^{2}-{\frac {3x}{5}}+{\frac {147}{25}}\right){\frac {-\mathrm {e} ^{-5x}}{5}}\right]\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7309e6f727f2209ba1a6e670ddf7f539cbaea918)
![{\displaystyle \left]k\pi -{\frac {\pi }{2}},k\pi +{\frac {\pi }{2}}\right[}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ed28e882190d9885e60372b85b0896744bfd5fb2)
![{\displaystyle \left]-\infty ,0\right[}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d10c3a4dbadb446d3509e7d7f25654de172d94be)
![{\displaystyle \left]0,+\infty \right[}](https://wikimedia.org/api/rest_v1/media/math/render/svg/010ebc9c2d0f3af5ce1c0594c44823f107968561)
![{\displaystyle \int f=\left[\mathrm {e} ^{x}\cos(2x)\right]+2\int g}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4765226932a5fe516bb63dba6427b8ec75ddf99c)
![{\displaystyle \int g=\left[\mathrm {e} ^{x}\sin(2x)\right]-2\int f}](https://wikimedia.org/api/rest_v1/media/math/render/svg/084fe0fa47e91f960580925e6a9ca07bf8a7a7bf)
![{\displaystyle {\begin{aligned}\int f&=\left[{\frac {\mathrm {e} ^{2x}}{2}}\sin 2x\right]-\int \mathrm {e} ^{2x}\cos 2x\,\mathrm {d} x\\&=\left[{\frac {\mathrm {e} ^{2x}}{2}}\left(\sin 2x-\cos 2x\right)\right]-\int f\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ee9f0deb4deace149817c89c02cb93619bef8d72)
![{\displaystyle {\begin{aligned}\int f&=\left[{\frac {\mathrm {e} ^{3x}}{3}}\left(\sin 2x-\cos 2x\right)\right]-{\frac {2}{3}}\int \mathrm {e} ^{3x}\left(\cos 2x+\sin 2x\right)\,\mathrm {d} x\\&=\left[{\frac {\mathrm {e} ^{3x}}{3}}\left(\sin 2x-\cos 2x-{\frac {2}{3}}\left(\cos 2x+\sin 2x\right)\right)\right]+{\frac {4}{9}}\int \mathrm {e} ^{3x}\left(-\sin 2x+\cos 2x\right)\,\mathrm {d} x\\&=\left[{\frac {\mathrm {e} ^{3x}\left(\sin 2x-5\cos 2x\right)}{9}}\right]-{\frac {4}{9}}\int f\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3bcb5116a63e835f0e8076ac093264323f280763)
![{\displaystyle \left]-1,+\infty \right[}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7cc6625e41160caeaf31e86f11c8fc9e3e4e943c)
![{\displaystyle \int f=\left[{\frac {x^{3}}{3}}\ln(x+1)\right]-{\frac {1}{3}}\int {\frac {x^{3}}{x+1}}\,\mathrm {d} x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e4efb6df1a458a0b005d3311d9f467d6477ab81b)
![{\displaystyle \int f=\left[x\cos \left(\ln x\right)\right]+\int g}](https://wikimedia.org/api/rest_v1/media/math/render/svg/73a621e475d880941022f78b758ce060dfdb11cb)
![{\displaystyle \int g=\left[x\sin \left(\ln x\right)\right]-\int f}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2aba4186d687fa572e5bd9476733e47870dbca0b)
![{\displaystyle \forall x>0\quad -\int _{x}^{+\infty }f(t)\;\mathrm {d} t=\left[{\frac {\ln(1+t^{2})}{t}}\right]_{x}^{+\infty }-\int _{x}^{+\infty }{\frac {2}{1+t^{2}}}\;\mathrm {d} t=\left[{\frac {\ln(1+t^{2})}{t}}-2\arctan t\right]_{x}^{+\infty }=-{\frac {\ln(1+x^{2})}{x}}-\pi +2\arctan x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4485ef1c94e2c1414b0a5d0a6fcf56215ce38b71)
![{\displaystyle \left]-\infty ,-{\frac {5}{3}}\right[}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b549941a1ead78fb0a43d1e27c275d3895e64e59)
![{\displaystyle \left]-{\frac {5}{3}},+\infty \right[}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d28064f3e6ace77dda871190e15ed84254777346)
![{\displaystyle \left[x\arctan x\right]-\int {\frac {x}{1+x^{2}}}\;\mathrm {d} x=x\arctan x-{\frac {\ln(1+x^{2})}{2}}+c}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8343651813c3f59d471b95bd0891d4d2dc2b6ac0)
![{\displaystyle \left[x\ln(x^{2}+2)\right]-2\int {\frac {x^{2}}{x^{2}+2}}\;\mathrm {d} x=\left[x\ln(x^{2}+2)\right]-2\int \left(1-{\frac {2}{x^{2}+2}}\right)\;\mathrm {d} x=x\ln(x^{2}+2)-2x+2{\sqrt {2}}\arctan {\frac {x}{\sqrt {2}}}+c}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0adaa6cdf13a24fd947e9c3a81d635e486377b37)
![{\displaystyle \left[{\frac {-\ln x}{1+x}}\right]+\int {\frac {1}{x(1+x)}}\;\mathrm {d} x=\left[{\frac {-\ln x}{1+x}}\right]+\int \left({\frac {1}{x}}-{\frac {1}{1+x}}\right)\;\mathrm {d} x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f8ccb2fef8641a8c2f41a70336fa755a8fe5ac33)
