# How to evaluate int_1^2(e^(2x)/(e^x-1)) ? Can you solve it without substitution and somehow use chain rule?

Jun 14, 2018

${\int}_{1}^{2} {e}^{2 x} / \left({e}^{x} - 1\right) \mathrm{dx} = e \left(e - 1\right) + \ln \left(e + 1\right)$

#### Explanation:

${\int}_{1}^{2} {e}^{2 x} / \left({e}^{x} - 1\right) \mathrm{dx} = {\int}_{1}^{2} {e}^{x} / \left({e}^{x} - 1\right) {e}^{x} \mathrm{dx}$

${\int}_{1}^{2} {e}^{2 x} / \left({e}^{x} - 1\right) \mathrm{dx} = {\int}_{1}^{2} {e}^{x} / \left({e}^{x} - 1\right) d \left({e}^{x}\right)$

${\int}_{1}^{2} {e}^{2 x} / \left({e}^{x} - 1\right) \mathrm{dx} = {\int}_{1}^{2} \frac{{e}^{x} - 1 + 1}{{e}^{x} - 1} d \left({e}^{x}\right)$

${\int}_{1}^{2} {e}^{2 x} / \left({e}^{x} - 1\right) \mathrm{dx} = {\int}_{1}^{2} \left(1 + \frac{1}{{e}^{x} - 1}\right) d \left({e}^{x}\right)$

${\int}_{1}^{2} {e}^{2 x} / \left({e}^{x} - 1\right) \mathrm{dx} = {\int}_{1}^{2} d \left({e}^{x}\right) + {\int}_{1}^{2} \frac{1}{{e}^{x} - 1} d \left({e}^{x}\right)$

${\int}_{1}^{2} {e}^{2 x} / \left({e}^{x} - 1\right) \mathrm{dx} = {\int}_{1}^{2} d \left({e}^{x}\right) + {\int}_{1}^{2} \frac{d \left({e}^{x} - 1\right)}{{e}^{x} - 1}$

${\int}_{1}^{2} {e}^{2 x} / \left({e}^{x} - 1\right) \mathrm{dx} = {\left[{e}^{x} + \ln \left({e}^{x} - 1\right)\right]}_{1}^{2}$

${\int}_{1}^{2} {e}^{2 x} / \left({e}^{x} - 1\right) \mathrm{dx} = {e}^{2} + \ln \left({e}^{2} - 1\right) - e - \ln \left(e - 1\right)$

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

${\int}_{1}^{2} {e}^{2 x} / \left({e}^{x} - 1\right) \mathrm{dx} = e \left(e - 1\right) + \ln \left(e + 1\right)$