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

解

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

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

解答ステップ

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

部分積分を適用する

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

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

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

簡素化

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

定数を解答に追加する

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