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