integral from 0 to 1 of (ln(x+1))/(x+1)

解

\int_{0}^{1} \frac{ln ( x + 1 )}{x + 1} d x

\frac{ln ^{2} ( 2 )}{2}

十進法表記

0.24022 \dots

解答ステップ

\int_{0}^{1} \frac{ln ( x + 1 )}{x + 1} d x

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

= \int_{0}^{1} ln (x + 1) \frac{1}{x + 1} d x

u置換積分法を適用する

= \int_{0}^{l n (2)} u d u

べき乗則を適用する

= [\frac{u ^{2}}{2}]_{0}^{l n (2)}

限度を計算する:

\frac{ln ^{2} ( 2 )}{2}

= \frac{ln ^{2} ( 2 )}{2}