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

解

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

- ln ∣ x - 1 ∣ + ln ∣ x ∣ - \frac{1}{x} + C

解答ステップ

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

以下の部分分数を得る:

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

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

\int \frac{1}{x - 1} d x = ln ∣ x - 1 ∣

\int \frac{1}{x} d x = ln ∣ x ∣

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

= - ln ∣ x - 1 ∣ + ln ∣ x ∣ - \frac{1}{x}

定数を解答に追加する

= - ln ∣ x - 1 ∣ + ln ∣ x ∣ - \frac{1}{x} + C