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

解

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

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

解答ステップ

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

拡張

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

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

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

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

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

定数を解答に追加する

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