Answer
$\frac{1}{48}tan^{-1}(\frac{3}{2})+\frac{1}{104}$
Work Step by Step
let
$x$ = $\frac{2}{3}tanθ$
$dx$ = $\frac{2}{3}sec^{2}θdθ$
and
$4+9x^{2}$ = $4+9(\frac{2}{3}tanθ)^{2}$ = $4+4tan^{2}θ$ = $4sec^{2}θ$
so
$\int{\frac{dx}{(4+9x^{2})^{2}}}$ = $\int{\frac{\frac{2}{3}sec^{2}θ}{16sec^{4}θ}dθ}$ = $\frac{1}{24}\int{cos^{2}θ}dθ$ = $\frac{1}{24}(\frac{1}{2}θ+\frac{1}{2}sinθcosθ+C$ = $\frac{1}{48}(θ+sinθ+cosθ)+C$
$x$ = $0$ then $θ$ = $0$
$x$ = $1$ then $tanθ$ = $\frac{3}{2}$
$sinθ$ = $\frac{3}{\sqrt {13}}$
$cosθ$ = $\frac{2}{\sqrt {13}}$
$\int_0^1{\frac{dx}{(4+9x^{2})^{2}}}$ = $\frac{1}{48}(θ+sinθ+cosθ)|_0^{tan^{-1}(\frac{3}{2})}$ = $\frac{1}{48}tan^{-1}(\frac{3}{2})+\frac{1}{104}$
