En posant ${\displaystyle x=\tan u}$, on obtient :

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

En posant ${\displaystyle x=\sinh u}$, on obtient :

${\displaystyle {\begin{aligned}\int _{0}^{\sqrt {3}}{\frac {x}{{\sqrt {1+x^{2}}}^{3}}}\,\mathrm {d} x&=\int _{0}^{\operatorname {arsinh} {\sqrt {3}}}{\frac {\sinh u}{{\sqrt {1+\sinh ^{2}u}}^{3}}}\,\cosh u\,\mathrm {d} u=\int _{0}^{\operatorname {arsinh} {\sqrt {3}}}{\frac {\sinh u}{\cosh ^{3}u}}\,\cosh u\,\mathrm {d} u\\&=\int _{0}^{\operatorname {arsinh} {\sqrt {3}}}{\frac {\sinh u}{\cosh ^{2}u}}\,\mathrm {d} u=\left[-{\frac {1}{\cosh u}}\right]_{0}^{\operatorname {arsinh} {\sqrt {3}}}=-{\frac {1}{\cosh \operatorname {arsinh} {\sqrt {3}}}}+{\frac {1}{\cosh 0}}\\&=-{\frac {1}{\sqrt {1+{\sqrt {3}}^{2}}}}+1=-{\frac {1}{2}}+1={\frac {1}{2}}.\end{aligned}}}$

${\displaystyle {\begin{aligned}\int _{\frac {1}{2}}^{2}{\frac {x-1}{(x+1){\sqrt {4x+5-x^{2}}}}}\,\mathrm {d} x&=\int _{\frac {1}{2}}^{2}{\frac {x-1}{(x+1){\sqrt {9-(x-2)^{2}}}}}\,\mathrm {d} x\\&=\int _{\frac {1}{2}}^{2}{\frac {x-1}{3(x+1){\sqrt {1-\left({\frac {x-2}{3}}\right)^{2}}}}}\,\mathrm {d} x.\end{aligned}}}$

On pose alors :

${\displaystyle \sin u={\frac {x-2}{3}}\Leftrightarrow x=3\sin u+2\qquad \mathrm {d} x=3\cos u\,\mathrm {d} u}$.

De plus, quand ${\displaystyle x}$ varie de ${\displaystyle {\frac {1}{2}}}$ à ${\displaystyle 2}$, ${\displaystyle \sin u}$ varie de ${\displaystyle -{\frac {1}{2}}}$ à ${\displaystyle 0}$ donc on peut faire varier ${\displaystyle u}$ de ${\displaystyle -{\frac {\pi }{6}}}$ à ${\displaystyle 0}$.

On a donc :

${\displaystyle {\begin{aligned}\int _{\frac {1}{2}}^{2}{\frac {x-1}{(x+1){\sqrt {4x+5-x^{2}}}}}\,\mathrm {d} x&=\int _{\frac {1}{2}}^{2}{\frac {x-1}{3(x+1){\sqrt {1-\left({\frac {x-2}{3}}\right)^{2}}}}}\,\mathrm {d} x\\&=\int _{-{\frac {\pi }{6}}}^{0}{\frac {3\sin u+2-1}{3(3\sin u+2+1){\sqrt {1-\sin ^{2}u}}}}3\cos u\,\mathrm {d} u\\&=\int _{-{\frac {\pi }{6}}}^{0}{\frac {1+3\sin u}{3(3+3\sin u)\cos u}}3\cos u\,\mathrm {d} u\\&={\frac {1}{3}}\int _{-{\frac {\pi }{6}}}^{0}{\frac {1+3\sin u}{1+\sin u}}\,\mathrm {d} u.\end{aligned}}}$

Aucune des règles de Bioche ne s’applique ici. Par conséquent, on pose :

${\displaystyle y=\tan {\frac {u}{2}}}$,

ce qui entraîne :

${\displaystyle \sin u={\frac {2y}{1+y^{2}}},\qquad \mathrm {d} u={\frac {2\,\mathrm {d} y}{1+y^{2}}}}$.

On a donc :

${\displaystyle {\begin{aligned}\int _{\frac {1}{2}}^{2}{\frac {x-1}{(x+1){\sqrt {4x+5-x^{2}}}}}\,\mathrm {d} x&={\frac {1}{3}}\int _{-{\frac {\pi }{6}}}^{0}{\frac {1+3\sin u}{1+\sin u}}\,\mathrm {d} u\\&={\frac {1}{3}}\int _{-\tan {\frac {\pi }{12}}}^{0}{\frac {3{\frac {2y}{1+y^{2}}}+1}{{\frac {2y}{1+y^{2}}}+1}}\,{\frac {2\,\mathrm {d} y}{1+y^{2}}}\\&={\frac {2}{3}}\int _{{\sqrt {3}}-2}^{0}{\frac {6y+1+y^{2}}{(2y+1+y^{2})(1+y^{2})}}\,\mathrm {d} y\\&={\frac {2}{3}}\int _{{\sqrt {3}}-2}^{0}{\frac {y^{2}+6y+1}{(y+1)^{2}(y^{2}+1)}}\,\mathrm {d} y,\end{aligned}}}$

qui se décompose en éléments simples sous la forme :

${\displaystyle {\begin{aligned}\int _{\frac {1}{2}}^{2}{\frac {x-1}{(x+1){\sqrt {4x+5-x^{2}}}}}\,\mathrm {d} x&=2\int _{{\sqrt {3}}-2}^{0}{\frac {1}{y^{2}+1}}\,\mathrm {d} y+{\frac {4}{3}}\int _{{\sqrt {3}}-2}^{0}{\frac {-1}{(y+1)^{2}}}\,\mathrm {d} y\\&=2\left[\arctan y\right]_{{\sqrt {3}}-2}^{0}+{\frac {4}{3}}\left[{\frac {1}{y+1}}\right]_{{\sqrt {3}}-2}^{0}\\&=2\times {\frac {\pi }{12}}+{\frac {4}{3}}\left(1-{\frac {1}{{\sqrt {3}}-1}}\right)\\&={\frac {\pi }{6}}+{\frac {2}{3}}-{\frac {2}{\sqrt {3}}}.\end{aligned}}}$

En posant

${\displaystyle x={\frac {1}{\cos u}},\qquad \mathrm {d} x={\frac {\sin u\mathrm {d} u}{\cos ^{2}u}}}$,

on vérifie que pour que ${\displaystyle x}$ varie de ${\displaystyle {\frac {2}{\sqrt {3}}}}$ à ${\displaystyle +\infty }$, on peut faire varier ${\displaystyle u}$ de ${\displaystyle {\frac {\pi }{6}}}$ à ${\displaystyle {\frac {\pi }{2}}}$.

On obtient :

${\displaystyle {\begin{aligned}\int _{\frac {2}{\sqrt {3}}}^{+\infty }{\frac {x^{2}}{{\sqrt {x^{2}-1}}^{5}}}\,\mathrm {d} x&=\int _{\frac {\pi }{6}}^{\frac {\pi }{2}}{\frac {1}{\cos ^{2}u{\sqrt {{\frac {1}{\cos ^{2}u}}-1}}^{5}}}\,{\frac {\sin u\,\mathrm {d} u}{\cos ^{2}u}}\\&=\int _{\frac {\pi }{6}}^{\frac {\pi }{2}}{\frac {\sin u\,\mathrm {d} u}{\cos ^{4}u\tan ^{5}u}}\\&=\int _{\frac {\pi }{6}}^{\frac {\pi }{2}}{\frac {\sin u\,\mathrm {d} u}{\cos ^{4}u{\frac {\sin ^{5}u}{\cos ^{5}u}}}}\\&=\int _{\frac {\pi }{6}}^{\frac {\pi }{2}}{\frac {\cos u\,\mathrm {d} u}{\sin ^{4}u}}\\&=\left[-{\frac {1}{3\sin ^{3}u}}\right]_{\frac {\pi }{6}}^{\frac {\pi }{2}}\\&=-{\frac {1}{3}}+{\frac {8}{3}}\\&={\frac {7}{3}}.\end{aligned}}}$

En posant

${\displaystyle x=\cosh u}$ on obtient, pour ${\displaystyle 1<a\leq b}$ :

${\displaystyle \int _{a}^{b}{\frac {\mathrm {d} x}{\sqrt {x^{2}-1}}}=\int _{\operatorname {arcosh} a}^{\operatorname {arcosh} b}{\frac {{\cancel {\sinh u}}\,\mathrm {d} u}{\cancel {\sinh u}}}}$

et l'on retrouve ainsi qu'une primitive sur ${\displaystyle \left]1,+\infty \right[}$ de ${\displaystyle x\mapsto {\frac {1}{\sqrt {x^{2}-1}}}}$ est la fonction arcosh.