# How do you integrate int (e^x+e^-x)/(e^x-e^-x)dx?

Feb 20, 2017

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

#### Explanation:

We know that

•coshx = (e^x + e^-x)/2
•sinhx = (e^x- e^-x)/2

This integral can be rewritten as

$\int \cosh \frac{x}{\sinh} x \mathrm{dx}$, where $\cosh x$ and $\sinh x$ represent the hyperbolic trigonometric functions

Now use a substitution to solve. Let $u = \sinh x$. Just like with regular trigonometric functions, $\mathrm{du} = \cosh x \mathrm{dx}$ and $\mathrm{dx} = \frac{\mathrm{du}}{\cosh} x$.

$\int \cosh \frac{x}{u} \cdot \frac{\mathrm{du}}{\cosh} x$

$\int \frac{1}{u} \mathrm{du}$

$\ln | u | + C$

$\ln | \sinh x | + C$

If you wish, the answer can be written as

$\ln | \frac{1}{2} \left({e}^{x} - {e}^{-} x\right) | + C$

Hopefully this helps!