Show that


$\int x^{2} e^{x^{3}} d x$ equals

(A) $\frac{1}{3} e^{x^{2}}+\mathrm{C}$

(B) $\frac{1}{3} e^{x^{2}}+\mathrm{C}$

(C) $\frac{1}{2} e^{x^{3}}+\mathrm{C}$

(D) $\frac{1}{3} e^{x^{2}}+\mathrm{C}$


Let $I=\int x^{2} e^{x^{3}} d x$

Also, let $x^{3}=t \Rightarrow 3 x^{2} d x=d t$

$\begin{aligned} \Rightarrow I &=\frac{1}{3} \int e^{t} d t \\ &=\frac{1}{3}\left(e^{t}\right)+\mathrm{C} \\ &=\frac{1}{3} e^{x^{3}}+\mathrm{C} \end{aligned}$

Hence, the correct answer is A.

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