Page:Annales de mathématiques pures et appliquées, 1830-1831, Tome 21.djvu/79

 Les corrections sont expliquées en page de discussion

$\int ^{m}\left\{(x+a)(x+b)\right\}^{r}\operatorname {d} x^{m}={\frac {1}{(r+m)^{m|-1}}}(x+a)^{r}(x+b)^{r+m}$

$-{\frac {m}{1}}{\frac {r}{(r+m+1)^{m+1|-1}}}(x+a)^{r-1}(x+b)^{r+m+1}$

$+{\frac {m}{1}}{\frac {m+1}{2}}{\frac {r^{2|-1}}{(r+m+2)^{m+2|-1}}}(x+a)^{r-2}(x+b)^{r+m+2}$

$-{\frac {m}{1}}.{\frac {m+1}{2}}.{\frac {m+2}{3}}{\frac {r^{3|-1}}{(r+m+3)^{m+3|-1}}}(x+a)^{r-3}(x+b)^{r+m+3}$

$\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \,;$(3)

ou bien encore

$\int ^{m}\left\{(x+a)(x+b)\right\}^{r}\operatorname {d} x^{m}={\frac {(x+a)^{r}(x+b)^{r+m}}{(r+m)^{m|-1}}}\times$

$\left\{1-{\frac {mr}{r+m+1}}\left({\frac {x+b}{x+a}}\right)+{\frac {m}{1}}{\frac {m+1}{2}}{\frac {r^{2|-1}}{(r+m+2)^{2|-1}}}\left({\frac {x+b}{x+a}}\right)^{2}\right.$

$\left.-{\frac {m}{1}}.{\frac {m+1}{2}}.{\frac {m+2}{3}}{\frac {r^{3|-1}}{(r+m+3)^{3|-1}}}\left({\frac {x+b}{x+a}}\right)^{3}+\ldots \right\}.$ (4)

Posant $m=1,$ il vient

$\int \left\{(x+a)(x+b)\right\}^{r}\operatorname {d} x={\frac {(x+a)^{r}(x+b)^{r+1}}{r+1}}\times$

$\left\{1-{\frac {r}{r+2}}\left({\frac {x+b}{x+a}}\right)+{\frac {r^{2|-1}}{(r+3)^{2|-1}}}\left({\frac {x+b}{x+a}}\right)^{2}-{\frac {r^{3|-1}}{(r+4)^{3|-1}}}\left({\frac {x+b}{x+a}}\right)^{3}\right.$

$\left.+{\frac {r^{4|-1}}{(r+5)^{4|-1}}}\left({\frac {x+b}{x+a}}\right)^{4}+\ldots \right\},$ (5)

Dans le cas particulier de $r=-{\frac {1}{2}},$ on trouve

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

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

ou bien encore