Answer
$$\frac{1}{6}\pi $$
Work Step by Step
$$\eqalign{
& \int_1^3 {\frac{{dx}}{{\sqrt x \left( {1 + x} \right)}}} \cr
& \int_1^3 {\frac{{dx}}{{\sqrt x \left( {1 + {{\left( {\sqrt x } \right)}^2}} \right)}}} \cr
& {\text{Let }}u = \sqrt x ,{\text{ }}du = \frac{1}{{2\sqrt x }}dx,{\text{ }}2du = \frac{1}{{\sqrt x }}dx \cr
& {\text{The new limits of integration are}} \cr
& x = 3 \Rightarrow u = \sqrt 3 \cr
& x = 1 \Rightarrow u = 1 \cr
& {\text{Substituting}} \cr
& \int_1^3 {\frac{{dx}}{{\sqrt x \left( {1 + {{\left( {\sqrt x } \right)}^2}} \right)}}} = \int_1^{\sqrt 3 } {\frac{{2du}}{{1 + {u^2}}}} \cr
& = 2\int_1^{\sqrt 3 } {\frac{{du}}{{1 + {u^2}}}} \cr
& {\text{Integrating}} \cr
& = 2\left[ {{{\tan }^{ - 1}}u} \right]_1^{\sqrt 3 } \cr
& = 2\left[ {{{\tan }^{ - 1}}\sqrt 3 - {{\tan }^{ - 1}}1} \right] \cr
& = 2\left( {\frac{1}{3}\pi - \frac{1}{4}\pi } \right) \cr
& = 2\left( {\frac{1}{{12}}\pi } \right) \cr
& = \frac{1}{6}\pi \cr} $$
