Answer
$$\frac{{{{\left( {{x^2} - 5x} \right)}^5}}}{5} + C$$
Work Step by Step
$$\eqalign{
& \int {{{\left( {{x^2} - 5x} \right)}^4}\left( {2x - 5} \right)} dx \cr
& {\text{set }}u = {x^2} - 5x{\text{ then }}\frac{{du}}{{dx}} = 2x - 5,\,\,\,\,\,\,\,\,\left( {2x - 5} \right)dx = du \cr
& {\text{write the integrand in terms of }}u \cr
& \int {{{\left( {{x^2} - 5x} \right)}^4}\left( {2x - 5} \right)} dx = \int {{u^4}du} \cr
& {\text{integrate by using the power rule }}\int {{u^n}} du = \frac{{{u^{n + 1}}}}{{n + 1}} + C \cr
& = \frac{{{u^{4 + 1}}}}{{4 + 1}} + C \cr
& = \frac{{{u^5}}}{5} + C \cr
& {\text{replace }}{x^2} - 5x{\text{ for }}u \cr
& = \frac{{{{\left( {{x^2} - 5x} \right)}^5}}}{5} + C \cr} $$
