Answer
$$\frac{1}{2}x\sqrt {9 + 16{x^2}} + \frac{9}{8}\ln \left| {4x + \sqrt {9 + 16{x^2}} } \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\sqrt {9 + 16{x^2}} dx} \cr
& {\text{Rewrite}} \cr
& \int {\sqrt {9 + 16{x^2}} dx} = \int {\sqrt {{{\left( 3 \right)}^2} + {{\left( {4x} \right)}^2}} dx} \cr
& {\text{Let }}u = 4x,{\text{ and }}a = 3,{\text{ }}du = 4dx,{\text{ }}dx = \frac{1}{4}du \cr
& {\text{Substitute}} \cr
& \int {\sqrt {{{\left( 3 \right)}^2} + {{\left( {4x} \right)}^2}} dx} = \int {\sqrt {{a^2} + {u^2}} \left( {\frac{1}{4}} \right)d} u \cr
& = \frac{1}{4}\int {\sqrt {{a^2} + {u^2}} d} u \cr
& {\text{Use the special integration formula }}\left( {{\text{Theorem 8}}{\text{.2}}} \right) \cr
& \int {\sqrt {{u^2} + {a^2}} du = \frac{1}{2}\left( {u\sqrt {{u^2} + {a^2}} + {a^2}\ln \left| {u + \sqrt {{u^2} + {a^2}} } \right|} \right) + C} \cr
& \frac{1}{4}\int {\sqrt {{a^2} + {u^2}} d} u = \frac{1}{8}\left( {u\sqrt {{u^2} + {a^2}} + {a^2}\ln \left| {u + \sqrt {{u^2} + {a^2}} } \right|} \right) + C \cr
& {\text{Substitute back}} \cr
& = \frac{1}{8}\left( {4x\sqrt {{{\left( 3 \right)}^2} + {{\left( {4x} \right)}^2}} + {{\left( 3 \right)}^2}\ln \left| {4x + \sqrt {{{\left( 3 \right)}^2} + {{\left( {4x} \right)}^2}} } \right|} \right) + C \cr
& = \frac{1}{2}x\sqrt {9 + 16{x^2}} + \frac{9}{8}\ln \left| {4x + \sqrt {9 + 16{x^2}} } \right| + C \cr} $$
