Question
Evalute : $\int \frac{e^x}{4 e^{2 x}-1} d x$

Answer

Let $I=\int \frac{e^x}{4 e^{2 x}-1} d x$
Put $e^x=t \quad \therefore e^x d x=d t$
$
\begin{aligned}
\therefore I & =\int \frac{1}{4 t^2-1} d t=\frac{1}{4} \int \frac{1}{ t ^2-\frac{1}{4}} d t \\
& =\frac{1}{4} \int \frac{1}{t^2-\left(\frac{1}{2}\right)^2} d t
\end{aligned}
$
$
\begin{aligned}
& =\frac{1}{4} \times \frac{1}{2 \times \frac{1}{2}} \log \left|\frac{t-\frac{1}{2}}{t+\frac{1}{2}}\right|+c \\
& =\frac{1}{4} \log \left|\frac{2 t-1}{2 t+1}\right|+c \\
& =\frac{1}{4} \log \left|\frac{2 e ^x-1}{2 e^x+1}\right|+c .
\end{aligned}
$

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