Answer
$\frac{x}{2}\sqrt {x^{2}-4}+2\ln|x+\sqrt {x^{2}-4}|+C$
Work Step by Step
substitution
$u$ = $x^{2}-4$
$du$ = $2xdx$
$I$ = $\int{\frac{x^{2}dx}{\sqrt {x^{2}-4}}}$ = $\int{\frac{u+4}{\sqrt u}}(\frac{du}{2\sqrt {u+4}})$ = $\frac{1}{2}\int{\frac{u+4}{\sqrt {u^{2}+4u}}}du$
so substitution is no effective for evaluating this integral
trigonometric
$x$ = $2secθ$
$dx$ = $3secθtanθdθ$
$I$ = $\int{\frac{x^{2}dx}{\sqrt {x^{2}-4}}}$ = $\int{\frac{4sec^{2}θ(2secθtanθ)dθ}{2tanθ}}$ = $4\int{sec^{3}θdθ}$ = $2tanθsecθ+2\ln|secθ+tanθ|+C$
$x$ = $2secθ$
$cosθ$ = $\frac{2}{x}$
$tanθ$ = $\frac{\sqrt {x^{2}-4}}{2}$
$I$ = $2tanθsecθ+2\ln|secθ+tanθ|+C$
$I$ = $2(\frac{1}{2}\sqrt {x^{2}-4})(\frac{x}{2})+2\ln|\frac{x}{2}+\frac{1}{2}\sqrt {x^{2}-4}|+C$
$I$ = $\frac{x}{2}\sqrt {x^{2}-4}+2\ln|\frac{1}{2}(x+\sqrt {x^{2}-4}|)+C$
$I$ = $\frac{x}{2}\sqrt {x^{2}-4}+2\ln|x+\sqrt {x^{2}-4}|+C$
