integral from 1 to 2 of x(ln(x))^2

解

\int_{1}^{2} x (ln (x))^{2} d x

2 ln^{2} (2) - 2 ln (2) + \frac{3}{4}

十進法表記

0.32461 \dots

解答ステップ

\int_{1}^{2} x (ln (x))^{2} d x

部分積分を適用する

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

\int x ln (x) d x = \frac{1}{2} x^{2} ln (x) - \frac{x ^{2}}{4}

= [\frac{1}{2} x^{2} ln^{2} (x) - (\frac{1}{2} x^{2} ln (x) - \frac{x ^{2}}{4})]_{1}^{2}

簡素化

= [\frac{1}{2} x^{2} ln^{2} (x) - \frac{1}{2} x^{2} ln (x) + \frac{x ^{2}}{4}]_{1}^{2}

限度を計算する:

2 ln^{2} (2) - 2 ln (2) + \frac{3}{4}

= 2 ln^{2} (2) - 2 ln (2) + \frac{3}{4}