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

解答

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

\frac{e ^{2}}{3} - 1

十进制

1.46301 \dots

求解步骤

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

使用分布积分法

= [- \frac{e ^{x} x}{1 + x} - \int - e^{x} d x]_{0}^{2}

\int - e^{x} d x = - e^{x}

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

化简

[- \frac{e ^{x} x}{1 + x} - (- e^{x})]_{0}^{2} : [\frac{e ^{x}}{1 + x}]_{0}^{2}

= [\frac{e ^{x}}{1 + x}]_{0}^{2}

带入上下限计算后得:

\frac{e ^{2}}{3} - 1

= \frac{e ^{2}}{3} - 1

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