Answer
$$\frac{\pi }{6}$$
Work Step by Step
$$\eqalign{
& \int_1^3 {\frac{{dx}}{{\sqrt x \left( {x + 1} \right)}}} dx \cr
& or \cr
& \int_1^3 {\frac{{dx}}{{\sqrt x \left( {{{\left( {\sqrt x } \right)}^2} + 1} \right)}}} dx \cr
& {\text{substitute }}u = \sqrt x ,{\text{ }}du = \frac{1}{{2\sqrt x }}dx \cr
& {\text{With this substitution}}{\text{,}} \cr
& {\text{if }}x = 3,{\text{ }}u = \sqrt 3 \cr
& {\text{if }}x = 3,{\text{ }}u = 1 \cr
& {\text{so}} \cr
& \int_1^3 {\frac{{dx}}{{\sqrt x \left( {{{\left( {\sqrt x } \right)}^2} + 1} \right)}}} dx = \int_1^{\sqrt 3 } {\frac{{2du}}{{{u^2} + 1}}} \cr
& {\text{find the antiderivative}} \cr
& = \left. {2\left( {{{\tan }^{ - 1}}u} \right)} \right|_1^{\sqrt 3 } \cr
& {\text{part 1 of fundamental theorem of calculus}} \cr
& = 2{\tan ^{ - 1}}\left( {\sqrt 3 } \right) - 2{\tan ^{ - 1}}\left( 1 \right) \cr
& {\text{simplify}} \cr
& = 2\left( {\frac{\pi }{3}} \right) - 2\left( {\frac{\pi }{4}} \right) \cr
& = \frac{{2\pi }}{3} - \frac{\pi }{2} \cr
& = \frac{\pi }{6} \cr} $$
