integral of (x^3)/((x^4-1)^2)

解

\int \frac{x ^{3}}{( x ^{4} - 1 ) ^{2}} d x

\frac{1}{8 ( x ^{2} + 1 )} + \frac{1}{16 ( x + 1 )} - \frac{1}{16 ( x - 1 )} + C

解答ステップ

\int \frac{x ^{3}}{( x ^{4} - 1 ) ^{2}} d x

以下の部分分数を得る:

\frac{x ^{3}}{( x ^{4} - 1 ) ^{2}} : - \frac{x}{4 ( x ^{2} + 1 ) ^{2}} - \frac{1}{16 ( x + 1 ) ^{2}} + \frac{1}{16 ( x - 1 ) ^{2}}

= \int - \frac{x}{4 ( x ^{2} + 1 ) ^{2}} - \frac{1}{16 ( x + 1 ) ^{2}} + \frac{1}{16 ( x - 1 ) ^{2}} d x

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int \frac{x}{4 ( x ^{2} + 1 ) ^{2}} d x - \int \frac{1}{16 ( x + 1 ) ^{2}} d x + \int \frac{1}{16 ( x - 1 ) ^{2}} d x

\int \frac{x}{4 ( x ^{2} + 1 ) ^{2}} d x = - \frac{1}{8 ( x ^{2} + 1 )}

\int \frac{1}{16 ( x + 1 ) ^{2}} d x = - \frac{1}{16 ( x + 1 )}

\int \frac{1}{16 ( x - 1 ) ^{2}} d x = - \frac{1}{16 ( x - 1 )}

= - (- \frac{1}{8 ( x ^{2} + 1 )}) - (- \frac{1}{16 ( x + 1 )}) - \frac{1}{16 ( x - 1 )}

簡素化

= \frac{1}{8 ( x ^{2} + 1 )} + \frac{1}{16 ( x + 1 )} - \frac{1}{16 ( x - 1 )}

定数を解答に追加する

= \frac{1}{8 ( x ^{2} + 1 )} + \frac{1}{16 ( x + 1 )} - \frac{1}{16 ( x - 1 )} + C