(int\frac{x^{2}}{e^{x^{3}}}dx =) \na (-\frac{1}{3}ln e^{x^{3}}+c) \nb (-\frac{e^{x^{3}}}{3}+c) \nc…

Explanation:

Step1: Use substitution

Let $u = x^{3}$, then $du=3x^{2}dx$, and $x^{2}dx=\frac{1}{3}du$.

Step2: Rewrite the integral

The original integral $\int\frac{x^{2}}{e^{x^{3}}}dx$ becomes $\frac{1}{3}\int\frac{1}{e^{u}}du=\frac{1}{3}\int e^{-u}du$.

Step3: Integrate $e^{-u}$

We know that $\int e^{-u}du=-e^{-u}+C$. So $\frac{1}{3}\int e^{-u}du =-\frac{1}{3}e^{-u}+C$.

Step4: Substitute back

Substitute $u = x^{3}$ back, we get $-\frac{1}{3e^{x^{3}}}+C$.

Answer:

C. $-\frac{1}{3e^{x^{3}}}+C$