Question
Evalute the following integrals: $\int\frac{\text{e}^{3\text{x}}}{\text{e}^{3\text{x}}+1}\text{dx}$

Answer

Let $\text{I}=\int\frac{\text{e}^{3\text{x}}}{\text{e}^{3\text{x}}+1}\text{dx}$ then,
Putting $e^{3x} + 1 = t$
$\Rightarrow3\text{e}^{3\text{x}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{dx}=\frac{\text{dt}}{3\text{e}^{3\text{x}}}$
$\therefore\text{I}=\int\frac{\text{e}^{3\text{x}}}{3\text{t}(\text{e}^{3\text{x}})}\text{dt}$
$=\frac{1}{3}\int\frac{1}{\text{t}}\text{dt}$
$=\frac{\text{In}|\text{t}|}{3}+\text{C}$
$=\frac{\text{In}|\text{e}^{3\text{x}}+1|}{3}+\text{C}$

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