Answer
$$\frac{1}{4}\ln \left| {\frac{{{e^t} - 2}}{{{e^t} + 2}}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{e^t}}}{{{e^{2t}} - 4}}dt} \cr
& {\text{Let }}x = {e^t},{\text{ }}dx = {e^t}dt \cr
& \int {\frac{{{e^t}}}{{{e^{2t}} - 4}}dt} = \int {\frac{{dx}}{{{x^2} - 4}}} \cr
& {\text{Decomposing }}\frac{1}{{{x^2} - 4}}{\text{ into partial fractions}} \cr
& \frac{1}{{{x^2} - 4}} = \frac{1}{{\left( {x - 2} \right)\left( {x + 2} \right)}} \cr
& \frac{1}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = \frac{A}{{x - 2}} + \frac{B}{{x + 2}} \cr
& 1 = A\left( {x + 2} \right) + B\left( {x - 2} \right) \cr
& {\text{let }}x = 2 \cr
& 1 = A\left( 4 \right),{\text{ }}A = \frac{1}{4} \cr
& {\text{let }}x = - 1 \cr
& 1 = B\left( { - 4} \right),{\text{ }}B = - \frac{1}{4} \cr
& \frac{1}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = \frac{{1/4}}{{x - 2}} - \frac{{1/4}}{{x + 2}} \cr
& ,{\text{ then}} \cr
& \int {\frac{{dx}}{{{x^2} - 4}}} = \int {\left( {\frac{{1/4}}{{x - 2}} - \frac{{1/4}}{{x + 2}}} \right)} dx \cr
& {\text{Integrating}} \cr
& = \frac{1}{4}\ln \left| {x - 2} \right| - \frac{1}{4}\ln \left| {x + 2} \right| + C \cr
& = \frac{1}{4}\ln \left| {\frac{{x - 2}}{{x + 2}}} \right| + C \cr
& {\text{Write in terms of }}\theta ,{\text{ replace }}x = {e^t} \cr
& = \frac{1}{4}\ln \left| {\frac{{{e^t} - 2}}{{{e^t} + 2}}} \right| + C \cr} $$
