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

解

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

\frac{1}{4 ( x + 1 ) ^{2}} - \frac{1}{4 ( x - 1 ) ^{2}} + C

解答ステップ

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

以下の部分分数を得る:

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

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

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

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

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

簡素化

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

定数を解答に追加する

= \frac{1}{4 ( x + 1 ) ^{2}} - \frac{1}{4 ( x - 1 ) ^{2}} + C