Answer
$$\ln \left| {1 + \sqrt x } \right| + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{2\sqrt x + 2x}}} \cr
& {\text{Factoring }}2\sqrt x + 2x,{\text{ the common factor is 2}}\sqrt x \cr
& = \int {\frac{{dx}}{{2\sqrt x \left( {1 + \sqrt x } \right)}}} \cr
& {\text{Use substitution:}}\cr
& {\text{Let }}u = 1 + \sqrt x,{\text{ so that }}du = \frac{1}{{2\sqrt x }}dx \cr
& dx = 2\sqrt x du \cr
& {\text{Write the integral in terms of }}u \cr
& \int {\frac{{dx}}{{2\sqrt x \left( {1 + \sqrt x } \right)}}} = \int {\frac{{2\sqrt x du}}{{2\sqrt x \left( u \right)}}} \cr
& = \int {\frac{1}{u}} du \cr
& {\text{Integrate }} \cr
& = \ln \left| u \right| + C \cr
& {\text{Write in terms of }}x:\cr
& {\text{Replace }}1 + \sqrt x \tan t{\text{ for }}u \cr
& = \ln \left| {1 + \sqrt x } \right| + C \cr} $$
