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 ex, 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 ex + e-x = t
Differentiating both sides, we get,
(ex - e-x )dx = dt
Now the integral becomes,
= $\int \frac{d t}{t}$ 
= log |t| + C
= log|ex + e-x| + C

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