Answer
$$8{\sin ^{ - 1}}\left( {\frac{{x - 3}}{3}} \right) - \sqrt {6x - {x^2}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{x + 5}}{{\sqrt {9 - {{\left( {x - 3} \right)}^2}} }}} dx \cr
& {\text{Let }}u = x - 3,{\text{ }}x = u + 3,{\text{ }}dx = du \cr
& {\text{Substitute}} \cr
& \int {\frac{{x + 5}}{{\sqrt {9 - {{\left( {x - 3} \right)}^2}} }}} dx = \int {\frac{{u + 3 + 5}}{{\sqrt {9 - {u^2}} }}} du \cr
& = \int {\frac{{u + 8}}{{\sqrt {9 - {u^2}} }}} du \cr
& = \int {\frac{u}{{\sqrt {9 - {u^2}} }}} du + \int {\frac{8}{{\sqrt {9 - {u^2}} }}} du \cr
& = - \frac{1}{2}\int {\frac{{ - 2u}}{{\sqrt {9 - {u^2}} }}} du + 8\int {\frac{1}{{\sqrt {{{\left( 3 \right)}^2} - {u^2}} }}} du \cr
& {\text{Integrate by the power rule and }}\int {\frac{{du}}{{\sqrt {{a^2} - {u^2}} }} = {{\sin }^{ - 1}}\left( {\frac{u}{a}} \right) + C} \cr
& = - \frac{1}{2}\left( {\frac{{\sqrt {9 - {u^2}} }}{{1/2}}} \right) + 8{\sin ^{ - 1}}\left( {\frac{u}{3}} \right) + C \cr
& = - \sqrt {9 - {u^2}} + 8{\sin ^{ - 1}}\left( {\frac{u}{3}} \right) + C \cr
& {\text{Write in terms of }}x \cr
& = - \sqrt {9 - {{\left( {x - 3} \right)}^2}} + 8{\sin ^{ - 1}}\left( {\frac{{x - 3}}{3}} \right) + C \cr
& = - \sqrt {9 - {x^2} + 6x - 9} + 8{\sin ^{ - 1}}\left( {\frac{{x - 3}}{3}} \right) + C \cr
& = 8{\sin ^{ - 1}}\left( {\frac{{x - 3}}{3}} \right) - \sqrt {6x - {x^2}} + C \cr} $$
