Answer
$$y = - \frac{1}{2}{x^2} - 4x + \frac{{10}}{3}\ln \left| {\frac{{x + 1}}{{x - 5}}} \right| + C$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} = \frac{{{x^3} - 21x}}{{5 + 4x - {x^2}}} \cr
& {\text{First use long division for }}\frac{{{x^3} - 21x}}{{5 + 4x - {x^2}}} \cr
& \frac{{dy}}{{dx}} = - x - 4 + \frac{{20}}{{5 + 4x - {x^2}}} \cr
& {\text{Completing the square}} \cr
& \frac{{dy}}{{dx}} = - x - 4 + \frac{{20}}{{9 - \left( {{x^2} - 4x + 4} \right)}} \cr
& \frac{{dy}}{{dx}} = - x - 4 + \frac{{20}}{{{{\left( 3 \right)}^2} - {{\left( {x - 2} \right)}^2}}} \cr
& {\text{Separate variables and integrate}} \cr
& dy = \left( { - x - 4 + \frac{{20}}{{{{\left( 3 \right)}^2} - {{\left( {x - 2} \right)}^2}}}} \right)dx \cr
& y = - \int {xdx} - \int {4dx} + 20\int {\frac{1}{{{{\left( 3 \right)}^2} - {{\left( {x - 2} \right)}^2}}}} dx \cr
& {\text{By theorem 5}}{\text{.20}} \cr
& y = - \frac{1}{2}{x^2} - 4x + 20\left( {\frac{1}{{2\left( 3 \right)}}} \right)\ln \left| {\frac{{3 + \left( {x - 2} \right)}}{{3 - \left( {x - 2} \right)}}} \right| + C \cr
& y = - \frac{1}{2}{x^2} - 4x + \frac{{10}}{3}\ln \left| {\frac{{x + 1}}{{x - 5}}} \right| + C \cr} $$
