Answer
$$\frac{1}{{48}}\left[ {\frac{2}{3} + \ln \left( {\sqrt 3 } \right)} \right]$$
Work Step by Step
$$\eqalign{
& \int_0^{1/3} {\frac{{dx}}{{{{\left( {4 - 9{x^2}} \right)}^2}}}} \cr
& {\text{write the integrand in terms of }}u \cr
& u = 3x,\,\,\,\,du = 3dx \cr
& \,\,\,\,x = 1/3 \to u = 1{\text{ and }}x = 0 \to u = 0 \cr
& {\text{then}} \cr
& \int_0^{1/3} {\frac{{dx}}{{{{\left( {4 - 9{x^2}} \right)}^2}}}} = \int_0^1 {\frac{{\left( {1/3} \right)du}}{{{{\left( {4 - {u^2}} \right)}^2}}}} \cr
& = \frac{1}{3}\int_0^1 {\frac{{du}}{{{{\left( {4 - {u^2}} \right)}^2}}}} \cr
& \cr
& {\text{Write the integrand in terms of }}\theta \cr
& u = 2\sin \theta ,\,\,\,du = 2\cos \theta d\theta ,\,\,\,\theta = {\sin ^{ - 1}}\left( {u/2} \right) \cr
& \,\,\,\,u = 1 \to \theta = \pi /6{\text{ and }}\,\,u = 0 \to \theta = 0 \cr
& \frac{1}{3}\int_0^1 {\frac{{du}}{{{{\left( {4 - {u^2}} \right)}^2}}}} = \frac{1}{3}\int_0^{\pi /6} {\frac{{2\cos \theta d\theta }}{{{{\left( {4 - {{\left( {2\sin \theta } \right)}^2}} \right)}^2}}}} \cr
& = \frac{2}{3}\int_0^{\pi /6} {\frac{{\cos \theta d\theta }}{{{{\left( {4 - 4{{\sin }^2}\theta } \right)}^2}}}} \cr
& = \frac{2}{3}\int_0^{\pi /6} {\frac{{\cos \theta }}{{16{{\cos }^4}\theta }}} d\theta \cr
& = \frac{1}{{24}}\int_0^{\pi /6} {\frac{1}{{{{\cos }^3}\theta }}} d\theta \cr
& = \frac{1}{{24}}\int_0^{\pi /6} {{{\sec }^3}\theta } d\theta \cr
& \cr
& {\text{Use the formula }}\int {{{\sec }^3}x} dx = \frac{1}{2}\sec x\tan x + \frac{1}{2}\ln \left| {\sec x + \tan x} \right| + C \cr
& \frac{1}{{24}}\int_0^{\pi /6} {{{\sec }^3}\theta } d\theta = \frac{1}{{24}}\left[ {\frac{1}{2}\sec \theta \tan \theta + \frac{1}{2}\ln \left| {\sec \theta + \tan \theta } \right|} \right]_0^{\pi /6} \cr
& = \frac{1}{{48}}\left[ {\sec \theta \tan \theta + \ln \left| {\sec \theta + \tan \theta } \right|} \right]_0^{\pi /6} \cr
& \cr
& {\text{evaluate the limits and simplify}} \cr
& = \frac{1}{{48}}\left[ {\sec \left( {\frac{\pi }{6}} \right)\tan \left( {\frac{\pi }{6}} \right) + \ln \left| {\sec \left( {\frac{\pi }{6}} \right) + \tan \left( {\frac{\pi }{6}} \right)} \right| - \sec 0\tan 0 - \ln \left| {\sec 0 + \tan 0} \right|} \right] \cr
& = \frac{1}{{48}}\left[ {\left( {\frac{{2\sqrt 3 }}{3}} \right)\left( {\frac{{\sqrt 3 }}{3}} \right) + \ln \left| {\frac{{2\sqrt 3 }}{3} + \frac{{\sqrt 3 }}{3}} \right| - 0 - 0} \right] \cr
& = \frac{1}{{48}}\left[ {\frac{2}{3} + \ln \left( {\sqrt 3 } \right)} \right] \cr} $$
