integral of 2/(x^3+x^2)

解

\int \frac{2}{x ^{3} + x ^{2}} d x

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

解答ステップ

\int \frac{2}{x ^{3} + x ^{2}} d x

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \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}

= 2 \cdot \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

= 2 (- \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 ∣

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

定数を解答に追加する

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