Answer
$$\frac{{x - 4}}{{\sqrt {9 + 8x - {x^2}} }} - {\sin ^{ - 1}}\left( {\frac{{x - 4}}{5}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{x^2} - 8x + 16}}{{{{\left( {9 + 8x - {x^2}} \right)}^{3/2}}}}} dx \cr
& {\text{Completing the square}} \cr
& 9 + 8x - {x^2} = 25 - {\left( {x - 4} \right)^2} \cr
& \int {\frac{{{{\left( {x - 4} \right)}^2}}}{{{{\left[ {25 - {{\left( {x - 4} \right)}^2}} \right]}^{3/2}}}}} dx \cr
& {\text{Let }}u = x - 4,\,\,\,du = dx \cr
& = \int {\frac{{{u^2}}}{{{{\left( {25 - {u^2}} \right)}^{3/2}}}}} du \cr
& {\text{The integrand contains the form }}{a^2} - {u^2} \cr
& 25 - {u^2} \to a = 5 \cr
& {\text{Use the change of variable }}u = a\sin \theta \cr
& = \int {\frac{{25{{\sin }^2}\theta }}{{{{\left( {25 - 25{{\sin }^2}\theta } \right)}^{3/2}}}}} \left( {5\cos \theta } \right)d\theta \cr
& = \int {\frac{{125{{\sin }^2}\theta }}{{125{{\left( {1 - {{\sin }^2}\theta } \right)}^{3/2}}}}} \left( {\cos \theta } \right)d\theta \cr
& = \int {\frac{{{{\sin }^2}\theta }}{{{{\left( {{{\cos }^2}\theta } \right)}^{3/2}}}}} \left( {\cos \theta } \right)d\theta \cr
& = \int {\frac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }}} d\theta \cr
& = \int {{{\tan }^2}\theta } d\theta \cr
& = \int {\left( {{{\sec }^2}\theta - 1} \right)} d\theta \cr
& {\text{Integrating}} \cr
& = \tan \theta - \theta + C \cr
& \cr
& {\text{Write in terms of }}u \cr
& = \frac{u}{{\sqrt {25 - {u^2}} }} - {\sin ^{ - 1}}\left( {\frac{u}{a}} \right) + C \cr
& {\text{Write in terms of }}x \cr
& = \frac{{x - 4}}{{\sqrt {9 + 8x - {x^2}} }} - {\sin ^{ - 1}}\left( {\frac{{x - 4}}{5}} \right) + C \cr} $$
