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

解

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

2 ln ∣ x - 1 ∣ + x + C

解答ステップ

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

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

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

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

\int 1 d x = x

= 2 ln ∣ x - 1 ∣ + x

定数を解答に追加する

= 2 ln ∣ x - 1 ∣ + x + C