Answer
$$\frac{x}{{16\sqrt {16 + 9{x^2}} }} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{{{\left( {16 + 9{x^2}} \right)}^{3/2}}}}} \cr
& u = 3x,{\text{ }}du = 3dx,{\text{ }}a = 4 \cr
& \int {\frac{{dx}}{{{{\left( {16 + 9{x^2}} \right)}^{3/2}}}}} = \frac{1}{3}\int {\frac{{du}}{{{{\left( {{a^2} + {u^2}} \right)}^{3/2}}}}} \cr
& {\text{a matching integral in a table of integrals at the end of the book is the }} \cr
& {\text{formula }}84 \cr
& \int {\frac{{du}}{{{{\left( {{a^2} + {u^2}} \right)}^{3/2}}}}} = \frac{u}{{{a^2}\sqrt {{a^2} + {u^2}} }} + C \cr
& \frac{1}{3}\int {\frac{{du}}{{{{\left( {{a^2} + {u^2}} \right)}^{3/2}}}}} = \frac{u}{{3{a^2}\sqrt {{a^2} + {u^2}} }} + C \cr
& u = 3x,{\text{ }}a = 4 \cr
& = \frac{{3x}}{{3{{\left( 4 \right)}^2}\sqrt {16 + 9{x^2}} }} + C \cr
& = \frac{x}{{16\sqrt {16 + 9{x^2}} }} + C \cr} $$
