Answer
$$\left( {12 + y} \right)\sqrt {36 + 3y} - \left( {3 + y} \right)\sqrt {9 + 3y} $$
Work Step by Step
$$\eqalign{
& \int_3^6 {x\sqrt {{x^2} + 3y} dx} \cr
& {\text{the notation }}dx{\text{ indicates integration with respect to }}x,{\text{ }} \cr
& {\text{so we treat }}x{\text{ as variable and }}y{\text{ as a constant}}{\text{. }} \cr
& {\text{write the integrand as follows}} \cr
& = \int_3^6 {x{{\left( {{x^2} + 3y} \right)}^{1/2}}dx} \cr
& = \frac{1}{2}\int_3^6 {{{\left( {{x^2} + 3y} \right)}^{1/2}}\left( {2x} \right)dx} \cr
& {\text{using the power rule }}\int {{u^n}} du = \frac{{{u^{n + 1}}}}{{n + 1}} + C \cr
& = \frac{1}{2}\left( {\frac{{{{\left( {{x^2} + 3y} \right)}^{3/2}}}}{{3/2}}} \right)_3^6 \cr
& = \frac{1}{2}\left( {\frac{{{{\left( {{x^2} + 3y} \right)}^{3/2}}}}{{3/2}}} \right)_3^6 \cr
& {\text{then}} \cr
& = \frac{1}{2}\left( {\frac{2}{3}} \right)\left( {{{\left( {{x^2} + 3y} \right)}^{3/2}}} \right)_3^6 \cr
& = \frac{1}{3}\left( {{{\left( {{x^2} + 3y} \right)}^{3/2}}} \right)_3^6 \cr
& = \frac{1}{3}\left( {\left( {{x^2} + 3y} \right)\sqrt {{x^2} + 3y} } \right)_3^6 \cr
& {\text{evaluating the limits in the variable }}x \cr
& = \frac{1}{3}\left[ {\left( {{{\left( 6 \right)}^2} + 3y} \right)\sqrt {{{\left( 6 \right)}^2} + 3y} } \right] - \frac{1}{3}\left[ {\left( {{{\left( 3 \right)}^2} + 3y} \right)\sqrt {{{\left( 3 \right)}^2} + 3y} } \right] \cr
& {\text{simplifying}} \cr
& = \frac{1}{3}\left[ {\left( {36 + 3y} \right)\sqrt {36 + 3y} } \right] - \frac{1}{3}\left[ {\left( {9 + 3y} \right)\sqrt {9 + 3y} } \right] \cr
& = \left( {12 + y} \right)\sqrt {36 + 3y} - \left( {3 + y} \right)\sqrt {9 + 3y} \cr} $$
