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受欢迎的微积分 >

integral of (e^{4x})/(1-e^{2x)}

解答

\int \frac{e ^{4 x}}{1 - e ^{2 x}} d x

解答

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

求解步骤

\int \frac{e ^{4 x}}{1 - e ^{2 x}} d x

因式分解

1 - e^{2 x} : - (- 1 + e^{2 x})

= \int \frac{e ^{4 x}}{- ( - 1 + e ^{2 x} )} d x

提出常数:

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

= - \int \frac{e ^{4 x}}{- 1 + e ^{2 x}} d x

使用换元积分法

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

乘开

\frac{u + 1}{u} : 1 + \frac{1}{u}

= - \frac{1}{2} \cdot \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

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

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= - \frac{1}{2} (u + ln ∣ u ∣)

u = - 1 + e^{2 x}

代回

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

解答补常数

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

作图

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\int (5 x - 1)^{2} d x

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\int x ln (x^{4} - 1) d x

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