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

解

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

- ln ∣ x - 1 ∣ + x + C

解答ステップ

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

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

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

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

\int 1 d x = x

= - ln ∣ x - 1 ∣ + x

定数を解答に追加する

= - ln ∣ x - 1 ∣ + x + C