Answer
$${e^x} - 2{\tan ^{ - 1}}\left( {\frac{{{e^x}}}{2}} \right) + C$$
Work Step by Step
$$\eqalign{
& {\text{Let }}u = {e^x},{\text{ }}du = {e^x}dx \cr
& \int {\frac{{{e^{3x}}}}{{{e^{2x}} + 4}}dx} = \int {\frac{{{u^2}}}{{{u^2} + 4}}du} \cr
& {\text{By the long division }}\frac{{{u^2}}}{{{u^2} + 4}} = 1 - \frac{1}{{{u^2} + 4}} \cr
& {\text{Rewrite the integrand}} \cr
& = \int {\left( {1 - \frac{1}{{{u^2} + 4}}} \right)du} \cr
& = \int {du} - \int {\frac{1}{{{u^2} + 4}}du} \cr
& {\text{Integrating}} \cr
& = u - 2{\tan ^{ - 1}}\left( {\frac{u}{2}} \right) + C \cr
& {\text{Write in terms of }}x,{\text{ replace }}u = {e^x} \cr
& = {e^x} - 2{\tan ^{ - 1}}\left( {\frac{{{e^x}}}{2}} \right) + C \cr} $$
