Answer
$$\frac{1}{8}\sqrt {16 + 8{x^2}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{x}{{\sqrt {16 + 8{x^2}} }}} dx \cr
& {\text{set }}u = 16 + 8{x^2}{\text{ then }}\frac{{du}}{{dx}} = 16x \to dx = \frac{{du}}{{16x}} \cr
& {\text{write the integrand in terms of }}u \cr
& \int {\frac{x}{{\sqrt {16 + 8{x^2}} }}} dx = \int {\frac{x}{{\sqrt u }}} \left( {\frac{{du}}{{16x}}} \right) \cr
& {\text{cancel the common factors}} \cr
& = \int {\frac{1}{{\sqrt u }}} \left( {\frac{{du}}{{16}}} \right) \cr
& = \frac{1}{{16}}\int {\frac{1}{{{u^{1/2}}}}} du \cr
& = \frac{1}{{16}}\int {{u^{ - 1/2}}} du \cr
& {\text{integrating by the powe rule }}\int {{u^n}} du = \frac{{{u^{n + 1}}}}{{n + 1}} + C \cr
& = \frac{1}{{16}}\left( {\frac{{{u^{1/2}}}}{{1/2}}} \right) + C \cr
& = \frac{2}{{16}}\sqrt u + C \cr
& = \frac{1}{8}\sqrt u + C \cr
& {\text{replace }}16 + 8{x^2}{\text{ for }}u \cr
& = \frac{1}{8}\sqrt {16 + 8{x^2}} + C \cr} $$
