integral of x 1/(x+1)

解

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

\frac{x ^{2}}{2 ( x + 1 )} + \frac{1}{2} x - ln ∣ x + 1 ∣ - \frac{1}{2 ( x + 1 )} + C

解答ステップ

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

部分積分を適用する

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

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

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

簡素化

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

定数を解答に追加する

= \frac{x ^{2}}{2 ( x + 1 )} + \frac{1}{2} x - ln ∣ x + 1 ∣ - \frac{1}{2 ( x + 1 )} + C