Answer
$$\int \frac{e^{x}}{4-e^{2x}}dx=\frac{1}{4}ln\left | \frac{e^{x}+2}{e^{x}-2} \right |+C$$
Work Step by Step
$\int \frac{du}{a^{2}-u^{2}}=\frac{1}{2a}ln\left | \frac{u+a}{u-a} \right |+C$
Thus:$$\int \frac{e^{x}}{4-e^{2x}}dx=\int \frac{d(e^{x})}{2^{2}-(e^{x})^{2}}$$
$$=\frac{1}{4}ln\left | \frac{e^{x}+2}{e^{x}-2} \right |+C$$
