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

解

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

- \frac{1}{32 2 ( x + 2 )} - \frac{1}{32 ( x + 2 ) ^{2}} + \frac{1}{32 2 ( x - 2 )} - \frac{1}{32 ( x - 2 ) ^{2}} + C

解答ステップ

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

以下の部分分数を得る:

\frac{x}{( x ^{2} - 2 ) ^{3}} : \frac{1}{32 2 ( x + 2 ) ^{2}} + \frac{1}{16 ( x + 2 ) ^{3}} - \frac{1}{32 2 ( x - 2 ) ^{2}} + \frac{1}{16 ( x - 2 ) ^{3}}

= \int \frac{1}{32 2 ( x + 2 ) ^{2}} + \frac{1}{16 ( x + 2 ) ^{3}} - \frac{1}{32 2 ( x - 2 ) ^{2}} + \frac{1}{16 ( x - 2 ) ^{3}} 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{1}{32 2 ( x + 2 ) ^{2}} d x + \int \frac{1}{16 ( x + 2 ) ^{3}} d x - \int \frac{1}{32 2 ( x - 2 ) ^{2}} d x + \int \frac{1}{16 ( x - 2 ) ^{3}} d x

\int \frac{1}{32 2 ( x + 2 ) ^{2}} d x = - \frac{1}{32 2 ( x + 2 )}

\int \frac{1}{16 ( x + 2 ) ^{3}} d x = - \frac{1}{32 ( x + 2 ) ^{2}}

\int \frac{1}{32 2 ( x - 2 ) ^{2}} d x = - \frac{1}{32 2 ( x - 2 )}

\int \frac{1}{16 ( x - 2 ) ^{3}} d x = - \frac{1}{32 ( x - 2 ) ^{2}}

= - \frac{1}{32 2 ( x + 2 )} - \frac{1}{32 ( x + 2 ) ^{2}} - (- \frac{1}{32 2 ( x - 2 )}) - \frac{1}{32 ( x - 2 ) ^{2}}

簡素化

= - \frac{1}{32 2 ( x + 2 )} - \frac{1}{32 ( x + 2 ) ^{2}} + \frac{1}{32 2 ( x - 2 )} - \frac{1}{32 ( x - 2 ) ^{2}}

定数を解答に追加する

= - \frac{1}{32 2 ( x + 2 )} - \frac{1}{32 ( x + 2 ) ^{2}} + \frac{1}{32 2 ( x - 2 )} - \frac{1}{32 ( x - 2 ) ^{2}} + C