integral of (xe^{-x^2})/(1+e^{-x^2)}

解答

\int \frac{x e ^{- x^{2}}}{1 + e ^{- x^{2}}} d x

- \frac{1}{2} ln 1 + e^{- x^{2}} + C

求解步骤

\int \frac{x e ^{- x^{2}}}{1 + e ^{- x^{2}}} d x

使用换元积分法

= \int - \frac{1}{2 u} d u

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \frac{1}{2} \cdot \int \frac{1}{u} d u

使用积分常见定则:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{2} ln ∣ u ∣

u = 1 + e^{- x^{2}}

代回

= - \frac{1}{2} ln 1 + e^{- x^{2}}

解答补常数

= - \frac{1}{2} ln 1 + e^{- x^{2}} + C

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