integral from 1 to 2 of x^4ln(x^2)

解

\int_{1}^{2} x^{4} ln (x^{2}) d x

- \frac{62}{25} + \frac{64}{5} ln (2)

十進法表記

6.39228 \dots

解答ステップ

\int_{1}^{2} x^{4} ln (x^{2}) d x

部分積分を適用する

= [\frac{1}{5} x^{5} ln (x^{2}) - \int \frac{2 x ^{4}}{5} d x]_{1}^{2}

\int \frac{2 x ^{4}}{5} d x = \frac{2 x ^{5}}{25}

= [\frac{1}{5} x^{5} ln (x^{2}) - \frac{2 x ^{5}}{25}]_{1}^{2}

\frac{1}{5} x^{5} ln (x^{2}) = \frac{2}{5} x^{5} ln (x)

= [\frac{2}{5} x^{5} ln (x) - \frac{2 x ^{5}}{25}]_{1}^{2}

限度を計算する:

- \frac{62}{25} + \frac{64}{5} ln (2)

= - \frac{62}{25} + \frac{64}{5} ln (2)