Answer
$$\frac{\pi }{{12}}$$
Work Step by Step
$$\eqalign{
& \int_0^1 {\frac{{dx}}{{2\sqrt {3 - x} \sqrt {x + 1} }}} \cr
& {\text{Let }}u = \sqrt {x + 1} ,{\text{ }}x = {u^2} - 1,{\text{ }}dx = 2udu \cr
& {\text{The new limits of integration are:}} \cr
& x = 0 \to u = 1 \cr
& x = 1 \to u = \sqrt 2 \cr
& {\text{Substituting}} \cr
& \int_0^1 {\frac{{dx}}{{2\sqrt {3 - x} \sqrt {x + 1} }}} = \int_1^{\sqrt 2 } {\frac{{2u}}{{2\sqrt {3 - \left( {{u^2} - 1} \right)} u}}} du \cr
& = \int_1^{\sqrt 2 } {\frac{1}{{\sqrt {4 - {u^2}} }}} du \cr
& {\text{Integrating}} \cr
& = \left[ {{{\sin }^{ - 1}}\left( {\frac{u}{2}} \right)} \right]_1^{\sqrt 2 } \cr
& = {\sin ^{ - 1}}\left( {\frac{{\sqrt 2 }}{2}} \right) - {\sin ^{ - 1}}\left( {\frac{1}{2}} \right) \cr
& {\text{Simplifying}} \cr
& = \frac{\pi }{4} - \frac{\pi }{6} \cr
& = \frac{\pi }{{12}} \cr} $$
