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

解

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

- \frac{3}{4} ln ∣ x + 1 ∣ + \frac{3}{4} ln ∣ x - 1 ∣ + \frac{1}{x + 1} - \frac{1}{2 ( x + 1 )} + C

解答ステップ

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

拡張

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

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

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

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

簡素化

- \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{x + 1} + \frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{2 ( x + 1 )} : - \frac{3}{4} ln ∣ x + 1 ∣ + \frac{3}{4} ln ∣ x - 1 ∣ + \frac{1}{x + 1} - \frac{1}{2 ( x + 1 )}

= - \frac{3}{4} ln ∣ x + 1 ∣ + \frac{3}{4} ln ∣ x - 1 ∣ + \frac{1}{x + 1} - \frac{1}{2 ( x + 1 )}

定数を解答に追加する

= - \frac{3}{4} ln ∣ x + 1 ∣ + \frac{3}{4} ln ∣ x - 1 ∣ + \frac{1}{x + 1} - \frac{1}{2 ( x + 1 )} + C