Question
$\int \frac{ e ^{3 x}}{ e ^{3 x}+1} d x$

Answer

Let $I =\int \frac{ e ^{3 x}}{ e ^{3 x}+1} d x$
Put $e ^{3 x }+1= t$
Differentiating w.r.t. $x$, we get
$ 3 e^{3 x} d x=d t$
$\therefore e^{3 x} d x=\frac{d t}{3}$
$\therefore I=\int \frac{1}{t} \cdot \frac{d t}{3}=\frac{1}{3} \log |t|+c$
$\therefore\left|\frac{1}{3} \log \right| e^{3 x}+1 \mid+c $

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