# What is int_(0)^(1) (e^(2x) - e^(-2x)) / (e^(2x) + e^(-2x))dx ?

Apr 9, 2018

Let $\left({e}^{2 x} + {e}^{- 2 x}\right) = t$

$\mathrm{dt} = \left[2 {e}^{2 x} - 2 {e}^{2 x}\right] \mathrm{dx}$

$\frac{\mathrm{dt}}{2} = \left[{e}^{2 x} - {e}^{- 2 x}\right] \mathrm{dx}$

The given integral is, ${\int}_{0}^{1} \frac{{e}^{2 x} - {e}^{- 2 x}}{{e}^{2 x} + {e}^{- 2 x}} \mathrm{dx}$

$\implies \int \frac{1}{t} \frac{\mathrm{dt}}{2}$

$\implies \frac{1}{2} \int \frac{\mathrm{dt}}{t}$

$\implies \frac{1}{2} \left[\log t\right]$

$\implies \frac{1}{2} {\left[\log \left({e}^{2 x} + {e}^{- 2 x}\right)\right]}_{0}^{1}$

$\implies \frac{1}{2} \left[\log \left({e}^{2} + \frac{1}{{e}^{2}}\right)\right] - \frac{1}{2} \log 2$

$\frac{1}{2} \left[\log \left({e}^{2} / 2 + \frac{1}{2 {e}^{2}}\right)\right]$

Apr 9, 2018

${\int}_{0}^{1} \setminus \frac{{e}^{2 x} - {e}^{- 2 x}}{{e}^{2 x} + {e}^{- 2 x}} \setminus \mathrm{dx} = \frac{1}{2} \setminus \ln \left(\frac{{e}^{2} + \frac{1}{e} ^ 2}{2}\right) \approx 0.66250$

#### Explanation:

We seek

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

Noting that:

$\sinh \left(x\right) = \frac{{e}^{x} - {e}^{- x}}{2}$ and $\cosh \left(x\right) = \frac{{e}^{x} + {e}^{- x}}{2}$

Then we can wrote:

$I = {\int}_{0}^{1} \setminus \sinh \frac{2 x}{\cosh} \left(2 x\right) \setminus \mathrm{dx}$

$\setminus \setminus = {\int}_{0}^{1} \setminus \tanh \left(2 x\right) \setminus \mathrm{dx}$

$\setminus \setminus = {\left[\frac{1}{2} \ln \left(\cosh 2 x\right)\right]}_{0}^{1}$

$\setminus \setminus = \frac{1}{2} \left\{\ln \left(\cosh 2\right) - \ln \left(\cosh 0\right)\right)$

$\setminus \setminus = \frac{1}{2} \left\{\ln \left(\frac{{e}^{2} + \frac{1}{e} ^ 2}{2}\right) - \ln \left(\frac{1 + 1}{2}\right)\right\}$

$\setminus \setminus = \frac{1}{2} \setminus \ln \left(\frac{{e}^{2} + \frac{1}{e} ^ 2}{2}\right)$

$\setminus \setminus \approx 0.66250$