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