Stuck on integrating $\int \frac{e^x}{e^{2x}-1}dx$

Question

I hope it's okay, but I wanted to show where I got stuck.

Problem: $$\int \frac{e^x}{e^{2x} -1} \ dx.$$

Attempt:

Let $u=e^{x}-1,$ and $u+1=e^x$ and $\ du= e^x \ dx.$

Then we get

$$\int \frac{u+1}{(u+1)^2-1} \ du.$$

Is there a possible trigonometric substitution of the form $$\int \frac{a}{x^2+a^2} \ dx$$ ?

Not sure.. have

$$\int \frac{e^x}{e^x \left(e^x-e^{-x} \right)} \ dx$$

Answer 1

Try letting $u=e^x.$ Then use partial fractions.

Answer 2

Alternatively, continuing from where you've left off: Note that $(u+1)^2 - 1 =u(u+2)$ via difference of squares and then apply partial fractions to $\frac{1}{u(u+2)}$. This isn't as efficient as Chris's answer but continues directly from your progress.

Note also that in your answer, you've forgotten to take into account the $\mathrm{d}u$ factor.

Answer 3

$$ I = \int \frac{e^{x}}{\color{blue}{e^{2x} - 1}} dx = \int \frac{e^{x}}{\left( \color{blue}{e^{x} + 1} \right)\color{blue}{\left( e^{x} - 1 \right)}} dx $$

Let $\color{red}{u=e^{x}+1}$, then $$ u=e^{x}+1, \quad du = e^{x} dx, \quad e^{x} - 1 = u - 2 $$

$$ I = \int \frac{du}{u(u-2)} = \frac{1}{2}\int \left( \frac{1}{u-2}-\frac{1}{u} \right) du = \frac{1}{2} \ln \left( \frac{2-u}{u} \right) $$

Substitute back $$ \int \frac{e^{x}}{e^{2x} - 1} dx = \frac{1}{2} \ln \left( \frac{1-e^{x}}{1+e^{x}} \right) $$

Answer 4

for a bit of variety make the substitution $x=2iz$ so the integral becomes: $$ \int \frac{e^{2iz}}{e^{4iz}-1}2idz = \int\frac{dz}{\sin 2z}=\frac12\int\frac{\sec^2 z}{\tan z}dz=\frac12 \log \tan z + C $$ since $$ \tan z = -i\frac{e^{iz}-e^{-iz}}{e^{iz}+e^{-iz}} = -i \frac{e^x-1}{e^x+1} $$ we have $$ I = \frac12\log \tan z + C = \frac12 \log \frac{e^x-1}{e^x+1} +C' $$ where $C'=C-\frac{i\pi}2$.