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

$\log \left({e}^{x} + {e}^{- x}\right) + C$
Note that $\frac{d}{\mathrm{dx}} \left({e}^{x} + {e}^{- x}\right) = {e}^{x} - {e}^{- x}$ so
$\frac{{e}^{x} - {e}^{- x}}{{e}^{x} + {e}^{- x}} = \frac{d}{\mathrm{dx}} \log \left({e}^{x} + {e}^{- x}\right)$ and finally
$\int \frac{{e}^{x} - {e}^{-} x}{{e}^{x} + {e}^{-} x} \mathrm{dx} = \int \frac{d}{\mathrm{dx}} \log \left({e}^{x} + {e}^{- x}\right) \mathrm{dx} = \log \left({e}^{x} + {e}^{- x}\right) + C$