What is the integral of (e^{3x})/(e^x-1) ?

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Expand

\frac{( u + 1 ) ^{2}}{u} : u + 2 + \frac{1}{u}

= \int u + 2 + \frac{1}{u} d u

Apply the Sum Rule:

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

= \int u d u + \int 2 d u + \int \frac{1}{u} d u

\int u d u = \frac{u ^{2}}{2}

\int 2 d u = 2 u

\int \frac{1}{u} d u = ln ∣ u ∣

= \frac{u ^{2}}{2} + 2 u + ln ∣ u ∣

Substitute back

u = e^{x} - 1

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

Add a constant to the solution

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

\frac{1}{ln ( x )}

\int x ln (8 x) d x

\int e^{4 x} sin (e^{4 x}) d x

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

\int q e^{q} + q d q

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