Difficult Integral, Need Help

Question

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

Answer 1

Hint. Use the substitution $y=e^x$.

If you really want to do it with the notions within the section, you could rewrite the integrand as follows:

$$\int\frac{e^x}{e^{2x}-1} dx = \int \frac{2e^{-x}}{2e^{-x}} \frac{e^x}{e^{2x}-1}\ dx = \frac{1}{2}\int\frac{2}{e^x - e^{-x}}dx = \frac{1}{2}\int\frac{1}{\sinh(x)}dx$$ which does not really simplify matters.

Answer 2

Just use substitution, where

$u=e^x$

$\int \frac{u}{u^2-1}\cdot \frac{1}{u}du$, this equals

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

$=-arctanh \left(e^x\right)+C$

Answer 3

Based on the background, you are looking at

$$\int\frac{e^x}{e^{2x}-1}\ dx=\int\frac1{e^x-e^{-x}}\ dx=\frac12\int\operatorname{csch}(x)\ dx$$

Intuition says

$$\int\operatorname{csch}(x)\ dx=\ln(\tanh\frac x2)+c$$

since,

$$\int\csc(x)\ dx=\ln(\tan\frac x2)+c$$

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Two hints:

$$I=\int\frac{e^x}{(e^x)^2-1}\ dx$$

Somehow, you'll want to end up with

$$\int\frac1{u^2-1}\ du$$

And PFD reveals

$$\frac1{u^2-1}=\frac12\left(\frac1{u-1}-\frac1{u+1}\right)$$