integral of (ln(x+1))/(x^2)

解

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

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

解答ステップ

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

部分積分を適用する

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

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

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

簡素化

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

定数を解答に追加する

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