Question
Evalute the following integrals: $\int\frac{\text{e}^{2\text{x}}}{\text{e}^{2\text{x}}-2}\text{dx}$

Answer

Let $\text{I}=\int\frac{\text{e}^{2\text{x}}}{\text{e}^{2\text{x}}-2}\text{dx}$
Putting $e^{2x} = t$
$\Rightarrow2\text{e}^{2\text{x}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{e}^{2\text{x}}\text{dx}=\frac{\text{dt}}{2}$
$\therefore\text{I}=\frac{1}{2}\int\frac{1}{\text{t}-2}\text{dt}$
$=\frac{1}{2}\text{ ln}|\text{t}-2|+\text{C}$
$=\frac{1}{2}\text{ ln}\big|\text{e}^{2\text{x}}-2\big|+\text{C }\big[\text{t}=\text{e}^{2\text{x}}\big]$

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