Answer
$\frac{1}{4}[\sqrt {17}-\sqrt 5]$
Work Step by Step
let
$x$ = $2tanθ$
$dx$ = $2sec^{2}θdθ$
and
$\sqrt {x^{2}+4}$ = $\sqrt {4tan^{2}θ+4}$ = $secθ$
so
$\int{\frac{dx}{x^{2}\sqrt {x^{2}+4}}}$ = $\int{\frac{2sec^{2}θdθ}{4tan^{2}θ(2secθ)}}$ = $\frac{1}{4}\int{\frac{cosθ}{sin^{2}θ}}dθ$
$u$ = $sinθ$
$du$ = $cosθdθ$
$\frac{1}{4}\int{\frac{cosθ}{sin^{2}θ}}dθ$ = $\frac{1}{4}\int{u^{-2}du}$ = $-\frac{1}{4sinθ}+C$
since
$x$ = $2tanθ$
$sinθ$ = $\frac{x}{\sqrt {x^{2}+4}}$
$\int_{\frac{1}{2}}^{1}{\frac{dx}{x^{2}\sqrt {x^{2}+4}}}$ = $-\frac{\sqrt {x^{2}+4}}{4x}|_{\frac{1}{2}}^{1}$ = $\frac{1}{4}[\sqrt {17}-\sqrt 5]$
