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}
et
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}
sous forme canonique. On obtient :
∫
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}}}}
et
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}}}