Answer
$\frac{1}{{\ln 2}}\ln \left| {{2^t} + 3} \right| + C$
Work Step by Step
$$\eqalign{
& \int {\frac{{{2^t}}}{{{2^t} + 3}}} dt \cr
& {\text{Let }}u = {2^t} + 3,{\text{ }}du = {2^t}\left( {\ln 2} \right)dt,{\text{ }}\frac{1}{{\ln 2}}du = {2^t}dt \cr
& {\text{Applying the substitution}} \cr
& \int {\frac{{{2^t}}}{{{2^t} + 3}}} dt = \int {\frac{1}{u}} \left( {\frac{1}{{\ln 2}}} \right)du \cr
& = \frac{1}{{\ln 2}}\int {\frac{1}{u}} du \cr
& {\text{Integrate }} \cr
& = \frac{1}{{\ln 2}}\ln \left| u \right| + C \cr
& {\text{Write in terms of }}t,{\text{ substitute }}u = {2^t} + 3 \cr
& = \frac{1}{{\ln 2}}\ln \left| {{2^t} + 3} \right| + C \cr} $$