Question
Integrate the function: $\frac{e^{2 x}~-1}{e^{2 x}~+1}$

Answer

We have, 
$\frac{e^{2 x}-1}{e^{2 x}+1}$
Dividing numerator and denominator by $e^x,$ we get,
$\frac{\frac{e^{2 x}-1}{e^{x}}}{\frac{e^{2 x}+1}{e^{-x}}}=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}$
Let $e^x + e^{-x} = t$
Differentiating both sides, we get,
$(e^x - e^{-x} )dx = dt$
Now the integral becomes,
$= \int \frac{d t}{t}$
$= \log |t| + C$
$= \log|e^x + e^{-x}| + C$

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