Answer
$$\frac{\pi }{{18}}$$
Work Step by Step
$$\eqalign{
& \int_0^{2/\sqrt 3 } {\frac{1}{{4 + 9{x^2}}}} dx \cr
& or \cr
& \int_0^{2/\sqrt 3 } {\frac{1}{{{{\left( 2 \right)}^2} + {{\left( {3x} \right)}^2}}}} dx \cr
& {\text{substitute }}u = 3x,{\text{ }}du = 3dx,{\text{ }} \cr
& {\text{With this substitution}}{\text{,}} \cr
& {\text{if }}x = 0,{\text{ }}u = 3\left( 0 \right) = 0 \cr
& {\text{if }}x = 2/\sqrt 3 ,{\text{ }}u = 3\left( {2/\sqrt 3 } \right) = 2\sqrt 3 \cr
& \int_0^{2/\sqrt 3 } {\frac{1}{{{{\left( 2 \right)}^2} + {{\left( {3x} \right)}^2}}}} dx = \frac{1}{3}\int_0^{2\sqrt 3 } {\frac{1}{{{2^2} + {u^2}}}} du \cr
& {\text{find the antiderivative }} \cr
& = \frac{1}{6}{\tan ^{ - 1}}\left. {\left( {\frac{u}{2}} \right)} \right|_0^{2\sqrt 3 } \cr
& {\text{part 1 of fundamental theorem of calculus}} \cr
& = \frac{1}{6}{\tan ^{ - 1}}\frac{{2\sqrt 3 }}{2} - \frac{1}{6}{\tan ^{ - 1}}\frac{0}{2} \cr
& {\text{simplifying}} \cr
& = \frac{1}{6}\left( {\frac{\pi }{3}} \right) \cr
& = \frac{\pi }{{18}} \cr} $$
