Prove
Question:

$\frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}}$

Solution:

Let $e^{2 x}+e^{-2 x}=t$

$\therefore\left(2 e^{2 x}-2 e^{-2 x}\right) d x=d t$

$\Rightarrow 2\left(e^{2 x}-e^{-2 x}\right) d x=d t$

$\Rightarrow \int\left(\frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}}\right) d x=\int \frac{d t}{2 t}$

$=\frac{1}{2} \int \frac{1}{t} d t$

$=\frac{1}{2} \log |t|+\mathrm{C}$

$=\frac{1}{2} \log \left|e^{2 x}+e^{-2 x}\right|+\mathrm{C}$

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