Page:Annales de mathématiques pures et appliquées, 1830-1831, Tome 21.djvu/80

$\int {\frac {\operatorname {d} x}{\sqrt {(x+a)(x+b)}}}=$

$2\left\{\left({\frac {x+a}{x+b}}\right)^{\frac {1}{2}}+{\frac {1}{3}}\left({\frac {x+a}{x+b}}\right)^{\frac {3}{2}}+{\frac {1}{5}}\left({\frac {x+b}{x+a}}\right)^{\frac {5}{2}}+{\frac {1}{7}}\left({\frac {x+b}{x+a}}\right)^{\frac {7}{2}}+\ldots \right\}.\mathrm {(6)}$

Or on a, dans le système Népérien,

$\operatorname {Log} .\left({\frac {1+u}{1-u}}\right)=2\left({\frac {u}{1}}+{\frac {u^{3}}{3}}+{\frac {u^{5}}{5}}+{\frac {u^{7}}{7}}+\ldots \right)\,;$

posant donc

${\frac {1+u}{1-u}}=z,\quad$d’où$\quad u={\frac {z-1}{z+1}}\,;$

on aura

$\operatorname {Log} .z=2\left\{\left({\frac {z-1}{z+1}}\right)+{\frac {1}{3}}\left({\frac {z-1}{z+1}}\right)^{3}+{\frac {1}{5}}\left({\frac {z-1}{z+1}}\right)^{5}+{\frac {1}{7}}\left({\frac {z-1}{z+1}}\right)^{7}+\ldots \right\}\,;$

donc, si l’on pose

$\left({\frac {x+a}{x+b}}\right)^{\frac {1}{2}}={\frac {z-1}{z+1}}\,;\qquad$(7)

on aura

$\int {\frac {\operatorname {d} x}{\sqrt {(x+a)(x+b)}}}=\operatorname {Log} .z\,;$

mais on tire de la formule (7)

$z={\frac {{\sqrt {x+b}}+{\sqrt {x+a}}}{{\sqrt {x+b}}-{\sqrt {x+a}}}}\,;$

donc finalement

$\int {\frac {\operatorname {d} x}{\sqrt {(x+a)(x+b)}}}=\operatorname {Log} .{\frac {{\sqrt {x+b}}+{\sqrt {x+a}}}{{\sqrt {x+b}}-{\sqrt {x+a}}}}.\qquad$(8)