# How do integrate this int (e^x + e^-x)/(e^x - e^-x)dx without hyperbolic indentities, preferebly with subsititution?

Jun 5, 2018

$= \ln \left({e}^{x} - {e}^{- x}\right) + C$

#### Explanation:

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

Substitute:

$u = {e}^{x} - {e}^{- x}$

From this it will follow that:

$\mathrm{du} = {e}^{x} + {e}^{- x} \mathrm{dx}$

So $\mathrm{du}$ is just the top of the fraction, now substituting in:

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

$= \ln \left(u\right) + C$

Reverse the substitution:

$= \ln \left({e}^{x} - {e}^{- x}\right) + C$