Calculus: Early Transcendentals (2nd Edition)

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