Une page de Wikiversité, la communauté pédagogique libre.
Début de la boite de navigation du chapitre
fin de la boite de navigation du chapitre
En raison de limitations techniques, la typographie souhaitable du titre, «
Changement de variable en calcul intégral : Intégrale contenant une racine n-ième d’une fonction homographique Changement de variable en calcul intégral/Intégrale contenant une racine n-ième d’une fonction homographique », n'a pu être restituée correctement ci-dessus.
Ces intégrales sont de la forme :
∫
α
β
f
(
x
,
a
x
+
b
c
x
+
d
n
)
d
x
{\displaystyle \int _{\alpha }^{\beta }f\left(x,\,{\sqrt[{n}]{\frac {ax+b}{cx+d}}}\right)\,\mathrm {d} x}
Traitons d'abord un cas simple :
n
=
2
,
c
=
0
{\displaystyle n=2,c=0}
Exemple : racine carrée d'un polynôme du premier degré [ modifier | modifier le wikicode ] Pour une intégrale de la forme :
∫
α
β
f
(
x
,
a
x
+
b
)
d
x
{\displaystyle \int _{\alpha }^{\beta }f\left(x,\,{\sqrt {ax+b}}\right)\,\mathrm {d} x}
on pose :
u
=
a
x
+
b
⇔
x
=
u
2
−
b
a
d
x
=
2
u
a
d
u
{\displaystyle u={\sqrt {ax+b}}\Leftrightarrow x={\frac {u^{2}-b}{a}}\qquad \mathrm {d} x={\frac {2u}{a}}\,\mathrm {d} u}
Ce changement de variable a permis de nous débarrasser de la racine gênante.
Exemple
Calculer :
∫
1
2
1
x
+
x
−
1
d
x
{\displaystyle \int _{1}^{2}{\frac {1}{x+{\sqrt {x-1}}}}\,\mathrm {d} x}
Posons :
u
=
x
−
1
⇔
x
=
u
2
+
1
d
x
=
2
u
d
u
{\displaystyle u={\sqrt {x-1}}\Leftrightarrow x=u^{2}+1\qquad \mathrm {d} x=2u\,\mathrm {d} u}
De plus :
x
=
1
⇔
u
=
0
{\displaystyle x=1\Leftrightarrow u=0}
x
=
2
⇔
u
=
1
{\displaystyle x=2\Leftrightarrow u=1}
Le calcul donne alors :
∫
1
2
1
x
+
x
−
1
d
x
=
∫
0
1
2
u
u
2
+
1
+
u
d
u
=
∫
0
1
2
u
+
1
u
2
+
u
+
1
d
u
−
∫
0
1
1
u
2
+
u
+
1
d
u
.
{\displaystyle {\begin{aligned}\int _{1}^{2}{\frac {1}{x+{\sqrt {x-1}}}}\,\mathrm {d} x&=\int _{0}^{1}{\frac {2u}{u^{2}+1+u}}\,\mathrm {d} u\\&=\int _{0}^{1}{\frac {2u+1}{u^{2}+u+1}}\,\mathrm {d} u-\int _{0}^{1}{\frac {1}{u^{2}+u+1}}\,\mathrm {d} u.\end{aligned}}}
La première intégrale ne pose pas de problème. Dans la deuxième, mettons
u
2
+
u
+
1
{\displaystyle u^{2}+u+1}
∫
1
2
1
x
+
x
−
1
d
x
=
[
ln
(
u
2
+
u
+
1
)
]
0
1
−
∫
0
1
1
(
u
+
1
2
)
2
+
3
4
d
u
=
ln
(
1
2
+
1
+
1
)
−
ln
(
0
2
+
0
+
1
)
−
4
3
∫
0
1
1
(
2
u
+
1
3
)
2
+
1
d
u
.
{\displaystyle {\begin{aligned}\int _{1}^{2}{\frac {1}{x+{\sqrt {x-1}}}}\,\mathrm {d} x&=\left[\ln(u^{2}+u+1)\right]_{0}^{1}-\int _{0}^{1}{\frac {1}{\left(u+{\frac {1}{2}}\right)^{2}+{\frac {3}{4}}}}\,\mathrm {d} u\\&=\ln(1^{2}+1+1)-\ln(0^{2}+0+1)-{\frac {4}{3}}\int _{0}^{1}{\frac {1}{\left({\frac {2u+1}{\sqrt {3}}}\right)^{2}+1}}\,\mathrm {d} u.\end{aligned}}}
Pour faire apparaître une dérivée de la fonction arc tangente, posons enfin :
t
=
2
u
+
1
3
⇔
u
=
t
3
−
1
2
d
u
=
3
2
d
t
{\displaystyle t={\frac {2u+1}{\sqrt {3}}}\Leftrightarrow u={\frac {t{\sqrt {3}}-1}{2}}\qquad \mathrm {d} u={\frac {\sqrt {3}}{2}}\,\mathrm {d} t}
On a aussi :
u
=
0
⇔
t
=
1
3
{\displaystyle u=0\Leftrightarrow t={\frac {1}{\sqrt {3}}}}
u
=
1
⇔
t
=
3
{\displaystyle u=1\Leftrightarrow t={\sqrt {3}}}
Par conséquent :
∫
1
2
1
x
+
x
−
1
d
x
=
ln
3
−
2
3
∫
1
3
3
1
t
2
+
1
d
t
=
ln
3
−
2
3
[
arctan
t
]
1
3
3
=
ln
3
−
2
3
(
arctan
3
−
arctan
1
3
)
=
ln
3
−
2
3
(
π
3
−
π
6
)
=
ln
3
−
π
3
3
.
{\displaystyle {\begin{aligned}\int _{1}^{2}{\frac {1}{x+{\sqrt {x-1}}}}\,\mathrm {d} x&=\ln 3-{\frac {2}{\sqrt {3}}}\int _{\frac {1}{\sqrt {3}}}^{\sqrt {3}}{\frac {1}{t^{2}+1}}\,\mathrm {d} t=\ln 3-{\frac {2}{\sqrt {3}}}\left[\arctan {t}\right]_{\frac {1}{\sqrt {3}}}^{\sqrt {3}}\\&=\ln 3-{\frac {2}{\sqrt {3}}}\left(\arctan {\sqrt {3}}-\arctan {\frac {1}{\sqrt {3}}}\right)=\ln 3-{\frac {2}{\sqrt {3}}}\left({\frac {\pi }{3}}-{\frac {\pi }{6}}\right)\\&=\ln 3-{\frac {\pi }{3{\sqrt {3}}}}.\end{aligned}}}
Dans le cas général d'une intégrale de la forme
∫
α
β
f
(
x
,
a
x
+
b
c
x
+
d
n
)
d
x
{\displaystyle \int _{\alpha }^{\beta }f\left(x,{\sqrt[{n}]{\frac {ax+b}{cx+d}}}\right)\,\mathrm {d} x}
on pose de même :
u
=
a
x
+
b
c
x
+
d
n
⇔
x
=
b
−
d
u
n
c
u
n
−
a
d
x
=
(
a
d
−
b
c
)
n
u
n
−
1
(
c
u
n
−
a
)
2
d
u
{\displaystyle u={\sqrt[{n}]{\frac {ax+b}{cx+d}}}\Leftrightarrow x={\frac {b-du^{n}}{cu^{n}-a}}\qquad \mathrm {d} x={\frac {(ad-bc)nu^{n-1}}{\left(cu^{n}-a\right)^{2}}}\,\mathrm {d} u}
Exemple
Calculer :
∫
−
1
2
0
1
2
x
−
1
1
+
x
1
−
x
d
x
{\displaystyle \int _{-{\frac {1}{2}}}^{0}{\frac {1}{2x-1}}{\sqrt {\frac {1+x}{1-x}}}\,\mathrm {d} x}
Posons
u
=
1
+
x
1
−
x
⇔
u
2
=
1
+
x
1
−
x
⇔
u
2
−
x
u
2
=
1
+
x
⇔
u
2
−
1
=
x
+
x
u
2
⇔
x
=
u
2
−
1
u
2
+
1
{\displaystyle u={\sqrt {\frac {1+x}{1-x}}}\Leftrightarrow u^{2}={\frac {1+x}{1-x}}\Leftrightarrow u^{2}-xu^{2}=1+x\Leftrightarrow u^{2}-1=x+xu^{2}\Leftrightarrow x={\frac {u^{2}-1}{u^{2}+1}}}
On a alors :
d
x
=
2
u
(
u
2
+
1
)
−
2
u
(
u
2
−
1
)
(
u
2
+
1
)
2
d
u
=
4
u
(
u
2
+
1
)
2
d
u
{\displaystyle \mathrm {d} x={\frac {2u(u^{2}+1)-2u(u^{2}-1)}{(u^{2}+1)^{2}}}\,\mathrm {d} u={\frac {4u}{(u^{2}+1)^{2}}}\,\mathrm {d} u}
On peut alors écrire :
∫
−
1
2
0
1
2
x
−
1
1
+
x
1
−
x
d
x
=
∫
1
3
1
u
2
+
1
u
2
−
3
4
u
2
(
u
2
+
1
)
2
d
u
=
∫
1
3
1
4
u
2
(
u
2
−
3
)
(
u
2
+
1
)
d
u
=
∫
1
3
1
3
u
2
−
3
d
u
+
∫
1
3
1
1
u
2
+
1
d
u
,
{\displaystyle {\begin{aligned}\int _{-{\frac {1}{2}}}^{0}{\frac {1}{2x-1}}{\sqrt {\frac {1+x}{1-x}}}\,\mathrm {d} x&=\int _{\frac {1}{\sqrt {3}}}^{1}{\frac {u^{2}+1}{u^{2}-3}}{\frac {4u^{2}}{(u^{2}+1)^{2}}}\,\mathrm {d} u=\int _{\frac {1}{\sqrt {3}}}^{1}{\frac {4u^{2}}{(u^{2}-3)(u^{2}+1)}}\,\mathrm {d} u\\&=\int _{\frac {1}{\sqrt {3}}}^{1}{\frac {3}{u^{2}-3}}\,\mathrm {d} u+\int _{\frac {1}{\sqrt {3}}}^{1}{\frac {1}{u^{2}+1}}\,\mathrm {d} u,\end{aligned}}}
qui se décompose en éléments simples sous la forme :
∫
−
1
2
0
1
2
x
−
1
1
+
x
1
−
x
d
x
=
3
2
∫
1
3
1
1
u
−
3
d
u
−
3
2
∫
1
3
1
1
u
+
3
d
u
+
∫
1
3
1
1
u
2
+
1
d
u
=
3
2
[
ln
|
u
−
3
|
−
ln
(
u
+
3
)
]
1
3
1
+
[
arctan
u
]
1
3
1
=
3
2
(
ln
3
−
1
1
+
3
−
ln
3
−
1
3
1
3
+
3
)
+
π
4
−
π
6
=
3
2
(
ln
(
3
−
1
)
2
2
−
ln
1
2
)
+
π
12
=
3
2
ln
(
(
3
−
1
)
2
)
+
π
12
=
3
ln
(
3
−
1
)
+
π
12
.
{\displaystyle {\begin{aligned}\int _{-{\frac {1}{2}}}^{0}{\frac {1}{2x-1}}{\sqrt {\frac {1+x}{1-x}}}\,\mathrm {d} x&={\frac {\sqrt {3}}{2}}\int _{\frac {1}{\sqrt {3}}}^{1}{\frac {1}{u-{\sqrt {3}}}}\,\mathrm {d} u-{\frac {\sqrt {3}}{2}}\int _{\frac {1}{\sqrt {3}}}^{1}{\frac {1}{u+{\sqrt {3}}}}\,\mathrm {d} u+\int _{\frac {1}{\sqrt {3}}}^{1}{\frac {1}{u^{2}+1}}\,\mathrm {d} u\\&={\frac {\sqrt {3}}{2}}\left[\ln \left|u-{\sqrt {3}}\right|-\ln \left(u+{\sqrt {3}}\right)\right]_{\frac {1}{\sqrt {3}}}^{1}+\left[\arctan u\right]_{\frac {1}{\sqrt {3}}}^{1}\\&={\frac {\sqrt {3}}{2}}\left(\ln {\frac {{\sqrt {3}}-1}{1+{\sqrt {3}}}}-\ln {\frac {{\sqrt {3}}-{\frac {1}{\sqrt {3}}}}{{\frac {1}{\sqrt {3}}}+{\sqrt {3}}}}\right)+{\frac {\pi }{4}}-{\frac {\pi }{6}}\\&={\frac {\sqrt {3}}{2}}\left(\ln {\frac {\left({\sqrt {3}}-1\right)^{2}}{2}}-\ln {\frac {1}{2}}\right)+{\frac {\pi }{12}}\\&={\frac {\sqrt {3}}{2}}\ln \left(\left({\sqrt {3}}-1\right)^{2}\right)+{\frac {\pi }{12}}\\&={\sqrt {3}}\,\ln \left({\sqrt {3}}-1\right)+{\frac {\pi }{12}}.\end{aligned}}}
![{\displaystyle \int _{\alpha }^{\beta }f\left(x,\,{\sqrt[{n}]{\frac {ax+b}{cx+d}}}\right)\,\mathrm {d} x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d92ccb3c6736fbd85c74babacd5cd96316aebe14)
![{\displaystyle {\begin{aligned}\int _{1}^{2}{\frac {1}{x+{\sqrt {x-1}}}}\,\mathrm {d} x&=\left[\ln(u^{2}+u+1)\right]_{0}^{1}-\int _{0}^{1}{\frac {1}{\left(u+{\frac {1}{2}}\right)^{2}+{\frac {3}{4}}}}\,\mathrm {d} u\\&=\ln(1^{2}+1+1)-\ln(0^{2}+0+1)-{\frac {4}{3}}\int _{0}^{1}{\frac {1}{\left({\frac {2u+1}{\sqrt {3}}}\right)^{2}+1}}\,\mathrm {d} u.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/38d5ad852c3f76630198eb9cb5e2056c5b9b69f7)
![{\displaystyle {\begin{aligned}\int _{1}^{2}{\frac {1}{x+{\sqrt {x-1}}}}\,\mathrm {d} x&=\ln 3-{\frac {2}{\sqrt {3}}}\int _{\frac {1}{\sqrt {3}}}^{\sqrt {3}}{\frac {1}{t^{2}+1}}\,\mathrm {d} t=\ln 3-{\frac {2}{\sqrt {3}}}\left[\arctan {t}\right]_{\frac {1}{\sqrt {3}}}^{\sqrt {3}}\\&=\ln 3-{\frac {2}{\sqrt {3}}}\left(\arctan {\sqrt {3}}-\arctan {\frac {1}{\sqrt {3}}}\right)=\ln 3-{\frac {2}{\sqrt {3}}}\left({\frac {\pi }{3}}-{\frac {\pi }{6}}\right)\\&=\ln 3-{\frac {\pi }{3{\sqrt {3}}}}.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d1ce44026070980093a2ef5e91da6913d3d2bc56)
![{\displaystyle \int _{\alpha }^{\beta }f\left(x,{\sqrt[{n}]{\frac {ax+b}{cx+d}}}\right)\,\mathrm {d} x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/187172016ffda3ad83b2990bca10d54d3a24fdf5)
![{\displaystyle u={\sqrt[{n}]{\frac {ax+b}{cx+d}}}\Leftrightarrow x={\frac {b-du^{n}}{cu^{n}-a}}\qquad \mathrm {d} x={\frac {(ad-bc)nu^{n-1}}{\left(cu^{n}-a\right)^{2}}}\,\mathrm {d} u}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0a28f86b7be2cfd00ffcdc32eeb64a7a3ad01681)
![{\displaystyle {\begin{aligned}\int _{-{\frac {1}{2}}}^{0}{\frac {1}{2x-1}}{\sqrt {\frac {1+x}{1-x}}}\,\mathrm {d} x&={\frac {\sqrt {3}}{2}}\int _{\frac {1}{\sqrt {3}}}^{1}{\frac {1}{u-{\sqrt {3}}}}\,\mathrm {d} u-{\frac {\sqrt {3}}{2}}\int _{\frac {1}{\sqrt {3}}}^{1}{\frac {1}{u+{\sqrt {3}}}}\,\mathrm {d} u+\int _{\frac {1}{\sqrt {3}}}^{1}{\frac {1}{u^{2}+1}}\,\mathrm {d} u\\&={\frac {\sqrt {3}}{2}}\left[\ln \left|u-{\sqrt {3}}\right|-\ln \left(u+{\sqrt {3}}\right)\right]_{\frac {1}{\sqrt {3}}}^{1}+\left[\arctan u\right]_{\frac {1}{\sqrt {3}}}^{1}\\&={\frac {\sqrt {3}}{2}}\left(\ln {\frac {{\sqrt {3}}-1}{1+{\sqrt {3}}}}-\ln {\frac {{\sqrt {3}}-{\frac {1}{\sqrt {3}}}}{{\frac {1}{\sqrt {3}}}+{\sqrt {3}}}}\right)+{\frac {\pi }{4}}-{\frac {\pi }{6}}\\&={\frac {\sqrt {3}}{2}}\left(\ln {\frac {\left({\sqrt {3}}-1\right)^{2}}{2}}-\ln {\frac {1}{2}}\right)+{\frac {\pi }{12}}\\&={\frac {\sqrt {3}}{2}}\ln \left(\left({\sqrt {3}}-1\right)^{2}\right)+{\frac {\pi }{12}}\\&={\sqrt {3}}\,\ln \left({\sqrt {3}}-1\right)+{\frac {\pi }{12}}.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2457cb23f3084f7bd231753cd4334f669440e507)
