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

解

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

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

解答ステップ

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

以下の部分分数を得る:

\frac{1}{x ^{2} ( 1 + x )} : - \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 ∣

定数を解答に追加する

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