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

解答

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

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

求解步骤

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

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 = e^{2 t} - 1

代回

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

解答补常数

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

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