Answer
$I$ = $\frac{1}{54}tan^{-1}(\frac{t}{3})+\frac{t}{18(t^{2}+9)}+C$
Work Step by Step
let
$t$ = $3tanθ$
$dt$ = $3sec^{2}θdθ$
$I$ = $\int{\frac{dt}{(t^{2}+9)^{2}}}$
$I$ = $\int{\frac{3sec^{2}θdθ}{81sec^{4}θ}}$
$I$ = $\frac{1}{27}\int{cos^{2}θdθ}$
$I$ = $\frac{1}{27}[\frac{1}{2}θ+\frac{1}{2}sinθcosθ]+C$
$t$ = $3tanθ$
$tanθ$ = $\frac{t}{3}$
$sinθ$ = $\frac{t}{\sqrt {t^{2}+9}}$
$cosθ$ = $\frac{3}{\sqrt {t^{2}+9}}$
$I$ = $\frac{1}{27}[\frac{1}{2}θ+\frac{1}{2}sinθcosθ]+C$
$I$ = $\frac{1}{54}tan^{-1}(\frac{t}{3})+\frac{1}{54}(\frac{t}{\sqrt {t^{2}+9}})(\frac{3}{\sqrt {t^{2}+9}})+C$
$I$ = $\frac{1}{54}tan^{-1}(\frac{t}{3})+\frac{t}{18(t^{2}+9)}+C$
