Answer
$$y = \frac{1}{4}{\sin ^{ - 1}}\left( {\frac{{4x - 1}}{9}} \right) + C$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} = \frac{1}{{\sqrt {80 + 8x - 16{x^2}} }} \cr
& {\text{Separating variables}} \cr
& dy = \frac{1}{{\sqrt {80 + 8x - 16{x^2}} }}dx \cr
& {\text{Completing the square}} \cr
& dy = \frac{1}{{\sqrt {81 - \left( {16{x^2} - 8x + 1} \right)} }}dx \cr
& dy = \frac{1}{{\sqrt {81 - {{\left( {4x - 1} \right)}^2}} }}dx \cr
& {\text{Let }}u = 4x - 1,{\text{ }}du = 4dx,{\text{ }}dx = \frac{1}{4}du \cr
& dy = \frac{1}{{\sqrt {81 - {u^2}} }}\left( {\frac{1}{4}} \right)du \cr
& {\text{Integrate both sides}} \cr
& \int {dy} = \frac{1}{4}\int {\frac{1}{{\sqrt {{{\left( 9 \right)}^2} - {u^2}} }}} du \cr
& y = \frac{1}{4}{\sin ^{ - 1}}\left( {\frac{u}{9}} \right) + C \cr
& {\text{Substitute }}u = 4x - 1 \cr
& y = \frac{1}{4}{\sin ^{ - 1}}\left( {\frac{{4x - 1}}{9}} \right) + C \cr} $$
