integral of (6x+6)/(x(x-1))

解

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

12 ln ∣ x - 1 ∣ - 6 ln ∣ x ∣ + C

解答ステップ

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

拡張

\frac{6 x + 6}{x ( x - 1 )} : \frac{6 x}{x ( x - 1 )} + \frac{6}{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{6 x}{x ( x - 1 )} d x + \int \frac{6}{x ( x - 1 )} d x

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

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

= 6 ln ∣ x - 1 ∣ + 6 (- ln ∣ x ∣ + ln ∣ x - 1 ∣)

簡素化

6 ln ∣ x - 1 ∣ + 6 (- ln ∣ x ∣ + ln ∣ x - 1 ∣) : 12 ln ∣ x - 1 ∣ - 6 ln ∣ x ∣

= 12 ln ∣ x - 1 ∣ - 6 ln ∣ x ∣

定数を解答に追加する

= 12 ln ∣ x - 1 ∣ - 6 ln ∣ x ∣ + C