第3章 3.11

(1)

$\frac{1}{x(x+1)} = \frac{1}{x} - \frac{1}{x+1}$

$f = \dfrac{1}{x} - \dfrac{1}{x+1}$ とおくと

$f' = -\frac{1}{x^2} + \frac{1}{(x+1)^2} \qquad f'' = \frac{2}{x^3} - \frac{2}{(x+1)^3}$

これから

$f^{(n)} = (-1)^n \frac{n!}{x^{n+1}} - (-1)^n \frac{n!}{(x+1)^{n+1}}$

(2)

$\frac{x}{x^2-1} = \frac{1}{2}\left(\frac{1}{x-1} + \frac{1}{x+1}\right)$

$f = \dfrac{1}{2}\left(\dfrac{1}{x-1} + \dfrac{1}{x+1}\right)$ とおくと

$f' = \frac{1}{2}\left\{\frac{-1}{(x-1)^2} + \frac{-1}{(x+1)^2}\right\}$

$f'' = \frac{1}{2}\left\{\frac{2}{(x-1)^3} + \frac{2}{(x+1)^3}\right\}$

これから

$f^{(n)} = (-1)^n \frac{n!}{2}\left\{\frac{1}{(x-1)^{n+1}} + \frac{1}{(x+1)^{n+1}}\right\}$

(3)

$\frac{x}{(x-1)(x+2)} = \frac{1}{3}\left(\frac{1}{x-1} + \frac{2}{x+2}\right)$

$f = \dfrac{1}{3}\left(\dfrac{1}{x-1} + \dfrac{2}{x+2}\right)$ とおくと

$f' = \frac{1}{3}\left\{\frac{-1}{(x-1)^2} + \frac{-2}{(x+2)^2}\right\}$

$f'' = \frac{1}{3}\left\{\frac{2}{(x-1)^3} + \frac{4}{(x+2)^3}\right\}$

これから

$f^{(n)} = (-1)^n \frac{n!}{3}\left\{\frac{1}{(x-1)^{n+1}} + \frac{2}{(x+2)^{n+1}}\right\}$

(4)

$\frac{x+1}{2x^2+x-3} = \frac{x+1}{(2x+3)(x-1)} = \frac{1}{5}\left(\frac{1}{2x+3} + \frac{2}{x-1}\right)$

$f = \dfrac{1}{5}\left(\dfrac{1}{2x+3} + \dfrac{2}{x-1}\right)$ とおくと

$f' = \frac{1}{5}\left\{\frac{-2}{(2x+3)^2} + \frac{-2}{(x-1)^2}\right\}$

$f'' = \frac{1}{5}\left\{\frac{8}{(2x+3)^3} + \frac{4}{(x-1)^3}\right\}$

これから

$f^{(n)} = (-1)^n \frac{n!}{5}\left\{\frac{2^n}{(2x+3)^{n+1}} + \frac{2}{(x-1)^{n+1}}\right\}$