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

解答

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

\frac{8 e ^{3} + 4}{9}

十进制

18.29825 \dots

求解步骤

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

使用分布积分法

= [\frac{1}{3} (2 x^{\frac{3}{2}} ln (x) - 3 \cdot \int \frac{2 x ^{\frac{1}{2}}}{3} d x)]_{1}^{e^{2}}

\int \frac{2 x ^{\frac{1}{2}}}{3} d x = \frac{4}{9} x^{\frac{3}{2}}

= [\frac{1}{3} (2 x^{\frac{3}{2}} ln (x) - 3 \cdot \frac{4}{9} x^{\frac{3}{2}})]_{1}^{e^{2}}

3 \cdot \frac{4}{9} x^{\frac{3}{2}} = \frac{4}{3} x^{\frac{3}{2}}

= [\frac{1}{3} (2 x^{\frac{3}{2}} ln (x) - \frac{4}{3} x^{\frac{3}{2}})]_{1}^{e^{2}}

带入上下限计算后得:

\frac{8 e ^{3}}{9} + \frac{4}{9}

= \frac{8 e ^{3}}{9} + \frac{4}{9}

化简

= \frac{8 e ^{3} + 4}{9}

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