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受欢迎的微积分 >

integral of (1/(e^x+e^{2x)})/(-e^{-3x)}

解答

\int \frac{\frac{1}{e ^{x} + e ^{2 x}}}{- e ^{- 3 x}} d x

解答

- 1 - e^{x} + ln ∣ 1 + e^{x} ∣ + C

求解步骤

\int \frac{\frac{1}{e ^{x} + e ^{2 x}}}{- e ^{- 3 x}} d x

提出常数:

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

= - \int \frac{\frac{1}{e ^{x} + e ^{2 x}}}{e ^{- 3 x}} d x

\frac{\frac{1}{e ^{x} + e ^{2 x}}}{e ^{- 3 x}} = \frac{1}{e ^{x} + e ^{2 x}} \cdot \frac{1}{e ^{- 3 x}}

= - \int \frac{1}{e ^{x} + e ^{2 x}} \cdot \frac{1}{e ^{- 3 x}} d x

化简

\frac{1}{e ^{x} + e ^{2 x}} \cdot \frac{1}{e ^{- 3 x}} : \frac{e ^{2 x}}{1 + e ^{x}}

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

使用换元积分法

= - \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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