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

解

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

3 ln ∣ x + 1 ∣ - 2 ln ∣ x ∣ + C

解答ステップ

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

拡張

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

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

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

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

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

簡素化

ln ∣ x + 1 ∣ - 2 (ln ∣ x ∣ - ln ∣ x + 1 ∣) : 3 ln ∣ x + 1 ∣ - 2 ln ∣ x ∣

= 3 ln ∣ x + 1 ∣ - 2 ln ∣ x ∣

定数を解答に追加する

= 3 ln ∣ x + 1 ∣ - 2 ln ∣ x ∣ + C