integral from 0 to e-1 of xln(x+1)

解答

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

\frac{e ^{2} - 3}{4}

十进制

1.09726 \dots

求解步骤

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

使用换元积分法

= \int_{1}^{e} (u - 1) ln (u) d u

使用分布积分法

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

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

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

化简

[ln (u) (\frac{u ^{2}}{2} - u) - (\frac{u ^{2}}{4} - u)]_{1}^{e} : [\frac{1}{4} (2 u^{2} ln (u) - 4 u ln (u) - u^{2} + 4 u)]_{1}^{e}

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

带入上下限计算后得:

\frac{e ^{2}}{4} - \frac{3}{4}

= \frac{e ^{2}}{4} - \frac{3}{4}

化简

= \frac{e ^{2} - 3}{4}

\int_{2}^{3} x^{3} - 4 x d x

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