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

解

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

\frac{1}{2}

十進法表記

0.5

解答ステップ

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

部分積分を適用する

= [\frac{1}{2} x^{2} ln (1 + \frac{1}{x}) - \int - \frac{x}{2 ( x + 1 )} d x]_{0}^{1}

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

= [\frac{1}{2} x^{2} ln (1 + \frac{1}{x}) - (- \frac{1}{2} (x + 1 - ln ∣ x + 1 ∣))]_{0}^{1}

簡素化

[\frac{1}{2} x^{2} ln (1 + \frac{1}{x}) - (- \frac{1}{2} (x + 1 - ln ∣ x + 1 ∣))]_{0}^{1} : [\frac{1}{2} (x^{2} ln (1 + \frac{1}{x}) + x + 1 - ln ∣ x + 1 ∣)]_{0}^{1}

= [\frac{1}{2} (x^{2} ln (1 + \frac{1}{x}) + x + 1 - ln ∣ x + 1 ∣)]_{0}^{1}

限度を計算する:

\frac{1}{2}

= \frac{1}{2}