Answer
$$\sqrt {4 + {x^2}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{xdx}}{{\sqrt {4 + {x^2}} }}} \cr
& \cr
& \left( {\bf{a}} \right){\text{using the substitution method}} \cr
& \,\,\,\,{\text{Let }}u = 4 + {x^2},\,\,\,\,\,du = 2xdx,\,\,\,\,dx = \frac{{du}}{{2x}} \cr
& \int {\frac{{xdx}}{{\sqrt {4 + {x^2}} }}} = \int {\frac{x}{{\sqrt u }}\left( {\frac{{du}}{{2x}}} \right)} \cr
& = \frac{1}{2}\int {\frac{1}{{\sqrt u }}du} = \frac{1}{2}\int {{u^{ - 1/2}}du} \cr
& {\text{integrate using the power rule}} \cr
& = \frac{1}{2}\left( {\frac{{{u^{1/2}}}}{{1/2}}} \right) + C \cr
& = \sqrt u + C \cr
& {\text{write in terms of }}x,{\text{ substitute }}4 + {x^2}{\text{ for }}u \cr
& = \sqrt {4 + {x^2}} + C \cr
& \cr
& \left( {\bf{b}} \right){\text{using a trigonometric substitution}} \cr
& \,\,\,\,\,{\text{Let }}x = 2\tan \theta ,\,\,\,dx = 2{\sec ^2}\theta d\theta \cr
& \,\,\,\,\,\int {\frac{{xdx}}{{\sqrt {4 + {x^2}} }}} = \int {\frac{{2\tan \theta \left( {2{{\sec }^2}\theta } \right)d\theta }}{{\sqrt {4 + {{\left( {2\tan \theta } \right)}^2}} }}} \cr
& = \int {\frac{{4\tan \theta {{\sec }^2}\theta d\theta }}{{\sqrt {4 + 4{{\tan }^2}\theta } }}} \cr
& = \int {\frac{{4\tan \theta {{\sec }^2}\theta d\theta }}{{2\sqrt {1 + {{\tan }^2}\theta } }}} \cr
& = \int {\frac{{2\tan \theta {{\sec }^2}\theta d\theta }}{{\sqrt {{{\sec }^2}\theta } }}} \cr
& = 2\int {\sec \theta \tan \theta } d\theta \cr
& {\text{Integrate}} \cr
& = 2\sec \theta + C \cr
& {\text{where sec}}\theta = \frac{{\sqrt {4 + {x^2}} }}{2} \cr
& = 2\left( {\frac{{\sqrt {4 + {x^2}} }}{2}} \right) + C \cr
& = \sqrt {4 + {x^2}} + C \cr} $$
