Answer
see below answer
Work Step by Step
direct substitution
let
$u$ = $x^{2}-4$
$du$ = $2xdx$
$I$ = $\int{\frac{xdx}{\sqrt {x^{2}-4}}}$ = $\frac{1}{2}\frac{du}{\sqrt u}$ = $\sqrt {u}+C$ = $\sqrt {x^{2}-4}+C$
trigonometric substitution
let
$x$ = $2secθ$
$dx$ = $2secθtanθdθ$
and
$x^{2}-4$ = $4sec^{2}θ-4$ = $4tan^{2}θ$
so
$I$ = $\int{\frac{xdx}{\sqrt {x^{2}-4}}}$ = $\int{\frac{2secθ(2secθtanθdθ)}{2tanθ}}$ = $2\int{sec^{2}θdθ}$ = $2tanθ+C$
$x$ = $2secθ$
$cosθ$ = $\frac{2}{x}$
$tanθ$ = $\frac{\sqrt {x^{2}-4}}{2}$
$I$ = $2tanθ+C$ = $2(\frac{\sqrt {x^{2}-4}}{2})+C$ = $\sqrt {x^{2}-4}+C$
