Answer
$$\frac{4}{3}{y^3} - 2{y^2} + y + C$$
Work Step by Step
$$\eqalign{
& \int {{{\left( {2y - 1} \right)}^2}} dy \cr
& {\text{expand the integrand using }}{\left( {A - B} \right)^2} = {A^2} - 2AB + {B^2} \cr
& = \int {\left( {4{y^2} - 4y + 1} \right)} dy \cr
& {\text{use sum rule for integrals}} \cr
& = \int {4{y^2}} dy - \int {4y} dy + \int {dy} \cr
& {\text{use }}\int {{y^n}dy} = \frac{{{y^{n + 1}}}}{{n + 1}} + C{\text{ and }}\int {dy} = y + C{\text{ then}} \cr
& = 4\left( {\frac{{{y^{2 + 1}}}}{{2 + 1}}} \right) - 4\left( {\frac{{{y^{1 + 1}}}}{{1 + 1}}} \right) + y + C \cr
& {\text{simplifying}} \cr
& = 4\left( {\frac{{{y^3}}}{3}} \right) - 4\left( {\frac{{{y^2}}}{2}} \right) + y + C \cr
& = \frac{4}{3}{y^3} - 2{y^2} + y + C \cr} $$
