Answer
$G\left( x \right) = \frac{4}{3}{x^{3/2}} - \frac{{2{x^{5/2}}}}{5} + \frac{{12{x^{7/2}}}}{7} + C$
Work Step by Step
$$\eqalign{
& g\left( x \right) = \sqrt x \left( {2 - x + 6{x^2}} \right) \cr
& {\text{First use the radical property }}\sqrt m = {m^{1/2}} \cr
& g\left( x \right) = {x^{1/2}}\left( {2 - x + 6{x^2}} \right) \cr
& {\text{Use the distributive property}} \cr
& g\left( x \right) = 2{x^{1/2}} - {x^{3/2}} + 6{x^{5/2}} \cr
& {\text{Using the formulas in Table 2 }}\left( {{\text{see page 358}}} \right){\text{ }} \cr
& {\text{Function: }}cf\left( x \right) \to {\text{Particular antiderivative: }}cF\left( x \right) \cr
& {\text{Function: }}{x^n}\left( {n \ne - 1} \right) \to {\text{Particular antiderivative: }}cF\left( x \right) \cr
& {\text{Function: }}f\left( x \right) + h\left( x \right) \to {\text{Particular antiderivative: }}f\left( x \right) + h\left( x \right) \cr
& {\text{And applying the Theorem 1}},{\text{ we obtain}} \cr
& G\left( x \right) = 2\left( {\frac{{{x^{1/2 + 1}}}}{{1/2 + 1}}} \right) - \frac{{{x^{3/2 + 1}}}}{{3/2 + 1}} + 6\left( {\frac{{{x^{5/2 + 1}}}}{{5/2 + 1}}} \right) + C \cr
& G\left( x \right) = 2\left( {\frac{{{x^{3/2}}}}{{3/2}}} \right) - \frac{{{x^{5/2}}}}{{5/2}} + 6\left( {\frac{{{x^{7/2}}}}{{7/2}}} \right) + C \cr
& G\left( x \right) = \frac{4}{3}{x^{3/2}} - \frac{{2{x^{5/2}}}}{5} + \frac{{12{x^{7/2}}}}{7} + C \cr} $$