Answer
$$6 - \frac{{3\pi }}{2}$$
Work Step by Step
$$\eqalign{
& \int_0^9 {\frac{{\sqrt x }}{{x + 9}}} dx \cr
& {\text{let }}u = \sqrt x ,\,\,\,\,{u^2} = x,\,\,\,\,2udu = dx \cr
& \int {\frac{{\sqrt x }}{{x + 9}}} dx = \int {\frac{u}{{{u^2} + 9}}\left( {2udu} \right)} \cr
& = \int {\frac{{2{u^2}}}{{{u^2} + 9}}} du \cr
& = \int {\left( {2 - \frac{{18}}{{{u^2} + 9}}} \right)} du \cr
& {\text{Integrating}} \cr
& = 2u - 18\left( {\frac{1}{3}{{\tan }^{ - 1}}\left( {\frac{u}{3}} \right)} \right) + C \cr
& = 2u - 6{\tan ^{ - 1}}\left( {\frac{u}{3}} \right) + C \cr
& {\text{substitute }}u = \sqrt x \cr
& = 2\sqrt x - 6{\tan ^{ - 1}}\left( {\frac{{\sqrt x }}{3}} \right) + C \cr
& {\text{Then}}{\text{,}} \cr
& \int_0^9 {\frac{{\sqrt x }}{{x + 9}}} dx = \left( {2\sqrt x - 6{{\tan }^{ - 1}}\left( {\frac{{\sqrt x }}{3}} \right)} \right)_0^9 \cr
& = \left( {2\sqrt 9 - 6{{\tan }^{ - 1}}\left( {\frac{{\sqrt 9 }}{3}} \right)} \right) - \left( {2\sqrt 0 - 6{{\tan }^{ - 1}}\left( {\frac{{\sqrt 0 }}{3}} \right)} \right) \cr
& = \left( {2\left( 3 \right) - 6{{\tan }^{ - 1}}\left( {\frac{3}{3}} \right)} \right) - \left( 0 \right) \cr
& = 6 - 6\left( {\frac{\pi }{4}} \right) \cr
& = 6 - \frac{{3\pi }}{2} \cr} $$
