Answer
\[ = 4\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{x\,\left( {3x + 2} \right)}}{{\sqrt {{x^3} + {x^2} + 4} }}dx} \hfill \\
\hfill \\
set\,\,\,\left( {{x^3} + {x^2} + 4} \right) = t{\text{ }}then\,\left( {3{x^2} + 2x} \right)dx = dt\,\,\,\,\, \hfill \\
\hfill \\
\to \,\,\,\,\,x\,\left( {3x + 2} \right)dx = dt \hfill \\
\hfill \\
integral\,\,becomes \hfill \\
\hfill \\
\int_{}^{} {\frac{{x\,\left( {3x + 2} \right)}}{{\sqrt {{x^3} + {x^2} + 4} }}dx} = \int_{}^{} {\frac{{dt}}{{\sqrt t }}} \hfill \\
\hfill \\
{\text{use }}\int {{x^n}dx = \frac{{{x^{n + 1}}}}{{n + 1}} + C} \hfill \\
\hfill \\
= 2\sqrt t + c \hfill \\
\hfill \\
substituting\,back\,\,t = {x^3} + {x^2} + 4 \hfill \\
\hfill \\
= 2\sqrt {{x^3} + {x^2} + 4} + C \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
\int_0^2 {\frac{{x\,\left( {3x + 2} \right)}}{{\sqrt {{x^3} + {x^2} + 4} }}dx} = 2\,\,\left[ {\sqrt {{x^3} + {x^2} + 4} } \right]_0^2 \hfill \\
\hfill \\
evaluate\,\,\,the\,\,{\text{limits}} \hfill \\
\hfill \\
= 2\,\left( {4 - 2} \right) = 4 \hfill \\
\hfill \\
\end{gathered} \]