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

解答

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

- 2 1 + e + ln (\frac{2 + 2 e + 1 + e}{2 + e - 2 e + 1}) - 4 (\frac{ln ( 1 + 2 ) - ln ( 2 - 1 )}{2} - 2)

十进制

0.57245 \dots

求解步骤

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

使用换元积分法

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

使用分布积分法

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

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

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

化简

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

带入上下限计算后得:

- 2 1 + e + 2 ln (e + 1 + 1) - 2 ln (e + 1 - 1) - 4 (\frac{ln ( 1 + 2 ) - ln ( 2 - 1 )}{2} - 2)

= - 2 1 + e + 2 ln (e + 1 + 1) - 2 ln (e + 1 - 1) - 4 (\frac{ln ( 1 + 2 ) - ln ( 2 - 1 )}{2} - 2)

化简

= - 2 1 + e + ln (\frac{2 + 2 e + 1 + e}{2 + e - 2 e + 1}) - 4 (\frac{ln ( 1 + 2 ) - ln ( 2 - 1 )}{2} - 2)

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