Answer
$$\frac{2}{{33}}\left( {\sqrt {7x + 297} - \sqrt {7x + 11} } \right)$$
Work Step by Step
$$\eqalign{
& \int_1^3 {\frac{{{y^2}}}{{\sqrt {7x + 11{y^3}} }}} dy \cr
& {\text{the notation }}dy{\text{ indicates integration with respect to }}y,{\text{ }} \cr
& {\text{so we treat }}y{\text{ as variable and }}x{\text{ as a constant}}{\text{. write the integrand as}} \cr
& = \int_1^3 {\frac{{{y^2}}}{{{{\left( {7x + 11{y^3}} \right)}^{1/2}}}}} dy \cr
& = \int_1^3 {{y^2}{{\left( {7x + 11{y^3}} \right)}^{ - 1/2}}} dy \cr
& = \frac{1}{{33}}\int_1^3 {{{\left( {7x + 11{y^3}} \right)}^{ - 1/2}}\left( {33{y^2}} \right)} dy \cr
& {\text{using the power rule }} \cr
& = \frac{1}{{33}}\left( {\frac{{{{\left( {7x + 11{y^3}} \right)}^{1/2}}}}{{1/2}}} \right)_1^3 \cr
& = \frac{2}{{33}}\left( {\sqrt {7x + 11{y^3}} } \right)_1^3 \cr
& {\text{evaluating the limits in the variable }}y \cr
& = \frac{2}{{33}}\left( {\sqrt {7x + 11{{\left( 3 \right)}^3}} - \sqrt {7x + 11{{\left( 1 \right)}^3}} } \right) \cr
& {\text{simplifying}} \cr
& = \frac{2}{{33}}\left( {\sqrt {7x + 297} - \sqrt {7x + 11} } \right) \cr} $$
