integral of (4x-2)/((x)(x+1)(x-2))

解

\int \frac{4 x - 2}{( x ) ( x + 1 ) ( x - 2 )} d x

ln ∣ x ∣ - 2 ln ∣ x + 1 ∣ + ln ∣ x - 2 ∣ + C

解答ステップ

\int \frac{4 x - 2}{x ( x + 1 ) ( x - 2 )} d x

以下の部分分数を得る:

\frac{4 x - 2}{x ( x + 1 ) ( x - 2 )} : \frac{1}{x} - \frac{2}{x + 1} + \frac{1}{x - 2}

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

= ln ∣ x ∣ - 2 ln ∣ x + 1 ∣ + ln ∣ x - 2 ∣

定数を解答に追加する

= ln ∣ x ∣ - 2 ln ∣ x + 1 ∣ + ln ∣ x - 2 ∣ + C