Question

1. (calculator not allowed)

\\(\int\frac{x^{2}}{e^{x^{3}}}dx =\\)
(a) \\(-\frac{1}{3}\ln e^{x^{3}}+c\\)
(b) \\(-\frac{e^{x^{3}}}{3}+c\\)
(c) \\(-\frac{1}{3e^{x^{3}}}+c\\)
(d) \\(\frac{1}{3}\ln e^{x^{3}}+c\\)
(e) \\(\frac{x^{3}}{3e^{x^{3}}}+c\\)

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 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 $u = x^{3}$

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

Answer:

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