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受欢迎的微积分 >

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

解答

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

解答

- 1 - e^{- x} + ln 1 + e^{- x} + C

求解步骤

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

使用换元积分法

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

提出常数:

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

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

乘开

\frac{u - 1}{u} : 1 - \frac{1}{u}

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

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= - (u - ln ∣ u ∣)

u = 1 + e^{- x}

代回

= - (1 + e^{- x} - ln 1 + e^{- x})

化简

= - 1 - e^{- x} + ln 1 + e^{- x}

解答补常数

= - 1 - e^{- x} + ln 1 + e^{- x} + C

作图

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n = 0 \sum \infty \frac{1}{n ^{2} + 3}

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