What is the integral of (e^{2x})/(sqrt(e^x-1)) ?

The integral of (e^{2x})/(sqrt(e^x-1)) is 2(1/3 (e^x-1)^{3/2}+sqrt(e^x-1))+C

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

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

2 (\frac{1}{3} (e^{x} - 1)^{\frac{3}{2}} + e^{x} - 1) + C

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

: \int 2 (v^{2} + 1) d v

= \int 2 (v^{2} + 1) d v

Take the constant out:

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

= 2 \cdot \int v^{2} + 1 d v

Apply the Sum Rule:

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

= 2 (\int v^{2} d v + \int 1 d v)

\int v^{2} d v = \frac{v ^{3}}{3}

\int 1 d v = v

= 2 (\frac{v ^{3}}{3} + v)

Substitute back

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

Simplify

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

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

Add a constant to the solution

= 2 (\frac{1}{3} (e^{x} - 1)^{\frac{3}{2}} + e^{x} - 1) + C

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What is the integral of (e^{2x})/(sqrt(e^x-1)) ?
The integral of (e^{2x})/(sqrt(e^x-1)) is 2(1/3 (e^x-1)^{3/2}+sqrt(e^x-1))+C