integral from 1 to e of x^5ln(x)

解答

\int_{1}^{e} x^{5} ln (x) d x

\frac{5 e ^{6} + 1}{36}

十进制

56.05955 \dots

求解步骤

\int_{1}^{e} x^{5} ln (x) d x

使用分布积分法

= [\frac{1}{6} (x^{6} ln (x) - 6 \cdot \int \frac{x ^{5}}{6} d x)]_{1}^{e}

\int \frac{x ^{5}}{6} d x = \frac{x ^{6}}{36}

= [\frac{1}{6} (x^{6} ln (x) - 6 \cdot \frac{x ^{6}}{36})]_{1}^{e}

6 \cdot \frac{x ^{6}}{36} = \frac{x ^{6}}{6}

= [\frac{1}{6} (x^{6} ln (x) - \frac{x ^{6}}{6})]_{1}^{e}

带入上下限计算后得:

\frac{5 e ^{6}}{36} + \frac{1}{36}

= \frac{5 e ^{6}}{36} + \frac{1}{36}

化简

= \frac{5 e ^{6} + 1}{36}

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