# int(e^(2x)+1)/(e^x-1)dx=?

## Apr 14, 2018

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

Taking ${e}^{x} = u$
$\implies \mathrm{du} = {e}^{x} \mathrm{dx}$

$\int \frac{{e}^{2 x} + 1}{{e}^{x} - 1} \mathrm{dx} = \int \frac{{u}^{2} + 1}{u - 1} \frac{\mathrm{du}}{e} ^ x$

$\implies \int \frac{{u}^{2} + 1}{u - 1} \times \frac{\mathrm{du}}{u}$

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

$\implies \int \frac{{u}^{2} + 1}{{u}^{2} - u} \mathrm{du}$

That's option E :)

Apr 14, 2018

$\int \setminus \frac{{e}^{2 x} + 1}{{e}^{x} - 1} \mathrm{dx} = {\left[\int \setminus \frac{{u}^{2} + 1}{{u}^{2} - u} \mathrm{du}\right]}_{u = {e}^{x}}$

#### Explanation:

First, we can rewrite the integral as follows:

$\int \setminus \frac{{e}^{2 x} + 1}{{e}^{x} - 1} \mathrm{dx} = \int \setminus \frac{{e}^{x} + \setminus \frac{1}{{e}^{x}}}{{e}^{x} - 1} {e}^{x} \mathrm{dx}$.

Written like this, we see that the integrand may be viewed as a composed function $f \left(g \left(x\right)\right)$ with the inner function $g \left(x\right) = g ' \left(x\right) = {e}^{x}$. Therefore,

$\int \setminus \frac{{e}^{x} + \setminus \frac{1}{{e}^{x}}}{{e}^{x} - 1} {e}^{x} \mathrm{dx} =$

$= {\left[\int \setminus \frac{u + \setminus \frac{1}{u}}{u - 1} \mathrm{du}\right]}_{u = {e}^{x}} =$

$= {\left[\int \setminus \frac{\setminus \frac{{u}^{2} + 1}{u}}{u - 1} \mathrm{du}\right]}_{u = {e}^{x}} =$

$= {\left[\int \setminus \frac{{u}^{2} + 1}{{u}^{2} - u} \mathrm{du}\right]}_{u = {e}^{x}}$.