# How to evaluate int_0^2(e^(2x)/(e^x-1))? Is there a way of not directly doing substitution?

Jun 13, 2018

${\int}_{0}^{2} \frac{{e}^{2 x} \cdot \mathrm{dx}}{{e}^{x} - 1} = \infty$

#### Explanation:

${\int}_{0}^{2} \frac{{e}^{2 x} \cdot \mathrm{dx}}{{e}^{x} - 1}$

=${\int}_{0}^{2} \frac{{e}^{x} \cdot {e}^{x} \cdot \mathrm{dx}}{{e}^{x} - 1}$

After using $u = {e}^{x} - 1$ and $\mathrm{du} = {e}^{x} \cdot \mathrm{dx}$ transforms,

${\int}_{0}^{{e}^{2} - 1} \frac{\left(u + 1\right) \cdot \mathrm{du}}{u}$

=$\infty$