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

解

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

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

解答ステップ

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

以下の部分分数を得る:

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

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

簡素化

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

定数を解答に追加する

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