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

解

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

- \frac{1}{2} ln ∣ x ∣ + \frac{1}{3} ln ∣ x + 1 ∣ + \frac{1}{6} ln ∣ x - 2 ∣ + C

解答ステップ

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

以下の部分分数を得る:

\frac{1}{x ( x + 1 ) ( x - 2 )} : - \frac{1}{2 x} + \frac{1}{3 ( x + 1 )} + \frac{1}{6 ( x - 2 )}

= \int - \frac{1}{2 x} + \frac{1}{3 ( x + 1 )} + \frac{1}{6 ( 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}{2 x} d x + \int \frac{1}{3 ( x + 1 )} d x + \int \frac{1}{6 ( x - 2 )} d x

\int \frac{1}{2 x} d x = \frac{1}{2} ln ∣ x ∣

\int \frac{1}{3 ( x + 1 )} d x = \frac{1}{3} ln ∣ x + 1 ∣

\int \frac{1}{6 ( x - 2 )} d x = \frac{1}{6} ln ∣ x - 2 ∣

= - \frac{1}{2} ln ∣ x ∣ + \frac{1}{3} ln ∣ x + 1 ∣ + \frac{1}{6} ln ∣ x - 2 ∣

定数を解答に追加する

= - \frac{1}{2} ln ∣ x ∣ + \frac{1}{3} ln ∣ x + 1 ∣ + \frac{1}{6} ln ∣ x - 2 ∣ + C