Answer
$$ = - \sqrt {5 + 4x - {x^2}} + 2{\sin ^{ - 1}}\left( {\frac{{x - 2}}{3}} \right) + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{x}{{\sqrt {5 + 4x - {x^2}} }}} dx \cr
& {\text{completing the square for the radicand}} \cr
& = - \left( {{x^2} - 4x - 5} \right) \cr
& = - \left( {{x^2} - 4x + 4 - 4 - 5} \right) \cr
& = - \left[ {{{\left( {x - 2} \right)}^2} - 9} \right] \cr
& = 9 - {\left( {x - 2} \right)^2} \cr
& \cr
& \int {\frac{x}{{\sqrt {5 + 4x - {x^2}} }}} dx = \int {\frac{x}{{\sqrt {9 - {{\left( {x - 2} \right)}^2}} }}} dx \cr
& {\text{make }}u = x - 2,\,\,\,\,{\text{then}}\,\,\,\,\,\,du = dx \cr
& \int {\frac{x}{{\sqrt {9 - {{\left( {x - 2} \right)}^2}} }}} dx = \int {\frac{{u + 2}}{{\sqrt {9 - {u^2}} }}} du \cr
& = \int {\frac{u}{{\sqrt {9 - {u^2}} }}} du + \int {\frac{2}{{\sqrt {9 - {u^2}} }}} du \cr
& = - \frac{1}{2}\int {\frac{{ - 2u}}{{\sqrt {9 - {u^2}} }}} du + \int {\frac{2}{{\sqrt {9 - {u^2}} }}} du \cr
& = - \sqrt {9 - {u^2}} + \int {\frac{2}{{\sqrt {9 - {u^2}} }}} du \cr
& \cr
& {\text{Use the Endpaper Integral Table to evaluate the integral }}\int {\frac{2}{{\sqrt {9 - {u^2}} }}} du \cr
& {\text{By formula 77}} \cr
& \left( {77} \right):\,\,\,\,\,\,\int {\frac{{du}}{{\sqrt {{a^2} - {u^2}} }}} = {\sin ^{ - 1}}\left( {\frac{u}{a}} \right) + C \cr
& \int {\frac{2}{{\sqrt {9 - {u^2}} }}} du = 2{\sin ^{ - 1}}\left( {\frac{u}{3}} \right) + C \cr
& \cr
& {\text{write in terms of }}x{\text{; replace }}x - 2{\text{ for }}u \cr
& = 2{\sin ^{ - 1}}\left( {\frac{{x - 2}}{3}} \right) + C \cr
& {\text{then}} \cr
& - \sqrt {9 - {u^2}} + \int {\frac{2}{{\sqrt {9 - {u^2}} }}} du = - \sqrt {9 - {{\left( {x - 2} \right)}^2}} + 2{\sin ^{ - 1}}\left( {\frac{{x - 2}}{3}} \right) + C \cr
& = - \sqrt {5 + 4x - {x^2}} + 2{\sin ^{ - 1}}\left( {\frac{{x - 2}}{3}} \right) + C \cr} $$
