Exemple

Calculer :

${\displaystyle \int _{1}^{2}{\frac {1}{x+{\sqrt {x-1}}}}\,\mathrm {d} x}$

Posons :

${\displaystyle u={\sqrt {x-1}}\Leftrightarrow x=u^{2}+1\qquad \mathrm {d} x=2u\,\mathrm {d} u}$

De plus :

${\displaystyle x=1\Leftrightarrow u=0}$ et ${\displaystyle x=2\Leftrightarrow u=1}$.

Le calcul donne alors :

${\displaystyle {\begin{aligned}\int _{1}^{2}{\frac {1}{x+{\sqrt {x-1}}}}\,\mathrm {d} x&=\int _{0}^{1}{\frac {2u}{u^{2}+1+u}}\,\mathrm {d} u\\&=\int _{0}^{1}{\frac {2u+1}{u^{2}+u+1}}\,\mathrm {d} u-\int _{0}^{1}{\frac {1}{u^{2}+u+1}}\,\mathrm {d} u.\end{aligned}}}$

La première intégrale ne pose pas de problème. Dans la deuxième, mettons ${\displaystyle u^{2}+u+1}$ sous forme canonique. On obtient :

${\displaystyle {\begin{aligned}\int _{1}^{2}{\frac {1}{x+{\sqrt {x-1}}}}\,\mathrm {d} x&=\left[\ln(u^{2}+u+1)\right]_{0}^{1}-\int _{0}^{1}{\frac {1}{\left(u+{\frac {1}{2}}\right)^{2}+{\frac {3}{4}}}}\,\mathrm {d} u\\&=\ln(1^{2}+1+1)-\ln(0^{2}+0+1)-{\frac {4}{3}}\int _{0}^{1}{\frac {1}{\left({\frac {2u+1}{\sqrt {3}}}\right)^{2}+1}}\,\mathrm {d} u.\end{aligned}}}$

Pour faire apparaître une dérivée de la fonction arc tangente, posons enfin :

${\displaystyle t={\frac {2u+1}{\sqrt {3}}}\Leftrightarrow u={\frac {t{\sqrt {3}}-1}{2}}\qquad \mathrm {d} u={\frac {\sqrt {3}}{2}}\,\mathrm {d} t}$.

On a aussi :

${\displaystyle u=0\Leftrightarrow t={\frac {1}{\sqrt {3}}}}$ et ${\displaystyle u=1\Leftrightarrow t={\sqrt {3}}}$.

Par conséquent :

${\displaystyle {\begin{aligned}\int _{1}^{2}{\frac {1}{x+{\sqrt {x-1}}}}\,\mathrm {d} x&=\ln 3-{\frac {2}{\sqrt {3}}}\int _{\frac {1}{\sqrt {3}}}^{\sqrt {3}}{\frac {1}{t^{2}+1}}\,\mathrm {d} t=\ln 3-{\frac {2}{\sqrt {3}}}\left[\arctan {t}\right]_{\frac {1}{\sqrt {3}}}^{\sqrt {3}}\\&=\ln 3-{\frac {2}{\sqrt {3}}}\left(\arctan {\sqrt {3}}-\arctan {\frac {1}{\sqrt {3}}}\right)=\ln 3-{\frac {2}{\sqrt {3}}}\left({\frac {\pi }{3}}-{\frac {\pi }{6}}\right)\\&=\ln 3-{\frac {\pi }{3{\sqrt {3}}}}.\end{aligned}}}$

Exemple

Calculer :

${\displaystyle \int _{-{\frac {1}{2}}}^{0}{\frac {1}{2x-1}}{\sqrt {\frac {1+x}{1-x}}}\,\mathrm {d} x}$.

Posons

${\displaystyle u={\sqrt {\frac {1+x}{1-x}}}\Leftrightarrow u^{2}={\frac {1+x}{1-x}}\Leftrightarrow u^{2}-xu^{2}=1+x\Leftrightarrow u^{2}-1=x+xu^{2}\Leftrightarrow x={\frac {u^{2}-1}{u^{2}+1}}}$.

On a alors :

${\displaystyle \mathrm {d} x={\frac {2u(u^{2}+1)-2u(u^{2}-1)}{(u^{2}+1)^{2}}}\,\mathrm {d} u={\frac {4u}{(u^{2}+1)^{2}}}\,\mathrm {d} u}$.

On peut alors écrire :

${\displaystyle {\begin{aligned}\int _{-{\frac {1}{2}}}^{0}{\frac {1}{2x-1}}{\sqrt {\frac {1+x}{1-x}}}\,\mathrm {d} x&=\int _{\frac {1}{\sqrt {3}}}^{1}{\frac {u^{2}+1}{u^{2}-3}}{\frac {4u^{2}}{(u^{2}+1)^{2}}}\,\mathrm {d} u=\int _{\frac {1}{\sqrt {3}}}^{1}{\frac {4u^{2}}{(u^{2}-3)(u^{2}+1)}}\,\mathrm {d} u\\&=\int _{\frac {1}{\sqrt {3}}}^{1}{\frac {3}{u^{2}-3}}\,\mathrm {d} u+\int _{\frac {1}{\sqrt {3}}}^{1}{\frac {1}{u^{2}+1}}\,\mathrm {d} u,\end{aligned}}}$

qui se décompose en éléments simples sous la forme :

${\displaystyle {\begin{aligned}\int _{-{\frac {1}{2}}}^{0}{\frac {1}{2x-1}}{\sqrt {\frac {1+x}{1-x}}}\,\mathrm {d} x&={\frac {\sqrt {3}}{2}}\int _{\frac {1}{\sqrt {3}}}^{1}{\frac {1}{u-{\sqrt {3}}}}\,\mathrm {d} u-{\frac {\sqrt {3}}{2}}\int _{\frac {1}{\sqrt {3}}}^{1}{\frac {1}{u+{\sqrt {3}}}}\,\mathrm {d} u+\int _{\frac {1}{\sqrt {3}}}^{1}{\frac {1}{u^{2}+1}}\,\mathrm {d} u\\&={\frac {\sqrt {3}}{2}}\left[\ln \left|u-{\sqrt {3}}\right|-\ln \left(u+{\sqrt {3}}\right)\right]_{\frac {1}{\sqrt {3}}}^{1}+\left[\arctan u\right]_{\frac {1}{\sqrt {3}}}^{1}\\&={\frac {\sqrt {3}}{2}}\left(\ln {\frac {{\sqrt {3}}-1}{1+{\sqrt {3}}}}-\ln {\frac {{\sqrt {3}}-{\frac {1}{\sqrt {3}}}}{{\frac {1}{\sqrt {3}}}+{\sqrt {3}}}}\right)+{\frac {\pi }{4}}-{\frac {\pi }{6}}\\&={\frac {\sqrt {3}}{2}}\left(\ln {\frac {\left({\sqrt {3}}-1\right)^{2}}{2}}-\ln {\frac {1}{2}}\right)+{\frac {\pi }{12}}\\&={\frac {\sqrt {3}}{2}}\ln \left(\left({\sqrt {3}}-1\right)^{2}\right)+{\frac {\pi }{12}}\\&={\sqrt {3}}\,\ln \left({\sqrt {3}}-1\right)+{\frac {\pi }{12}}.\end{aligned}}}$