integral from 0 to 2 of x^2ln(x^2+1)

解

\int_{0}^{2} x^{2} ln (x^{2} + 1) d x

\frac{8}{3} ln (5) - \frac{2 ( 3 arctan ( 2 ) + 2 )}{9}

十進法表記

3.10929 \dots

解答ステップ

\int_{0}^{2} x^{2} ln (x^{2} + 1) d x

部分積分を適用する

= [\frac{1}{3} x^{3} ln (x^{2} + 1) - \int \frac{2 x ^{4}}{3 ( x ^{2} + 1 )} d x]_{0}^{2}

\int \frac{2 x ^{4}}{3 ( x ^{2} + 1 )} d x = \frac{2}{3} (\frac{x ^{3}}{3} - x + arctan (x))

= [\frac{1}{3} x^{3} ln (x^{2} + 1) - \frac{2}{3} (\frac{x ^{3}}{3} - x + arctan (x))]_{0}^{2}

限度を計算する:

\frac{8}{3} ln (5) - \frac{2 ( 3 arctan ( 2 ) + 2 )}{9}

= \frac{8}{3} ln (5) - \frac{2 ( 3 arctan ( 2 ) + 2 )}{9}