Answer
$$\frac{1}{3}\sqrt {11} + \sqrt 3 $$
Work Step by Step
$$\eqalign{
& \int_0^4 {\frac{x}{{\sqrt {3 + 2x} }}} dx \cr
& {\text{Let }}u = 2x,{\text{ }}du = 2dx,{\text{ }}dx = \frac{1}{2}du \cr
& {\text{The new limits of integration are}} \cr
& x = 4 \to u = 8 \cr
& x = 0 \to u = 0 \cr
& {\text{Substituting}} \cr
& \int_0^4 {\frac{x}{{\sqrt {3 + 2x} }}} dx = \frac{1}{8}\int_0^8 {\frac{{u/2}}{{\sqrt {3 + u} }}} \left( {\frac{1}{2}} \right)du \cr
& = \frac{1}{4}\int_0^8 {\frac{u}{{\sqrt {3 + u} }}} du \cr
& {\text{Integrating by tables }}\int {\frac{u}{{\sqrt {a + bu} }}} du = \frac{{ - 2\left( {2a - bu} \right)}}{{3{b^2}}}\sqrt {a + bu} + C \cr
& \frac{1}{4}\int_0^8 {\frac{u}{{\sqrt {3 + u} }}} du = \frac{1}{4}\left[ {\frac{{ - 2\left( {2\left( 3 \right) - u} \right)}}{3}\sqrt {3 + u} } \right]_0^8 \cr
& = - \frac{1}{6}\left[ {\left( {6 - u} \right)\sqrt {3 + u} } \right]_0^8 \cr
& = - \frac{1}{6}\left[ {\left( {6 - 8} \right)\sqrt {3 + 8} } \right] + \frac{1}{6}\left[ {\left( {6 - 0} \right)\sqrt {3 + 0} } \right] \cr
& {\text{Simplifying}} \cr
& = \frac{1}{3}\sqrt {11} + \sqrt 3 \cr} $$
