Answer
$$2\sqrt {x - 2} - 2\sqrt 3 {\tan ^{ - 1}}\sqrt {\frac{{x - 2}}{3}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\sqrt {x - 2} }}{{x + 1}}} dx \cr
& {\text{Let }}u = \sqrt {x - 2} ,{\text{ }}{u^2} = x - 2,{\text{ }}2udu = dx \cr
& {\text{Substituting}} \cr
& \int {\frac{{\sqrt {x - 2} }}{{x + 1}}} dx = \int {\frac{u}{{{u^2} + 2 + 1}}} \left( {2udu} \right) \cr
& = \int {\frac{{2{u^2}}}{{{u^2} + 3}}} du \cr
& {\text{By long division }}\frac{{2{u^2}}}{{{u^2} + 3}} = 2 - \frac{6}{{{u^2} + 3}} \cr
& = \int {\left( {2 - \frac{6}{{{u^2} + 3}}} \right)} du \cr
& {\text{Integrating}} \cr
& = 2u - 6\left( {\frac{1}{{\sqrt 3 }}{{\tan }^{ - 1}}\left( {\frac{u}{{\sqrt 3 }}} \right)} \right) + C \cr
& = 2u - \frac{6}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{u}{{\sqrt 3 }}} \right) + C \cr
& {\text{Write in terms of }}x \cr
& = 2\sqrt {x - 2} - \frac{6}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{{\sqrt {x - 2} }}{{\sqrt 3 }}} \right) + C \cr
& = 2\sqrt {x - 2} - 2\sqrt 3 {\tan ^{ - 1}}\sqrt {\frac{{x - 2}}{3}} + C \cr} $$
