Question

\int \frac { x ^ { 2 } } { e ^ { x ^ { 3 } } } d x =

a

• \frac { 1 } { 3 } \ln e ^ { x ^ { 3 } } + c

b

• \frac { e ^ { x ^ { 3 } } } { 3 } + c

c

• \frac { 1 } { 3 e ^ { x ^ { 3 } } } + c

d
\frac { 1 } { 3 } \ln e ^ { x ^ { 3 } } + c

e
\frac { x ^ { 3 } } { 3 e ^ { x ^ { 3 } } } + c

Explanation:

Step1: Simplify the integrand

We know that \(\frac{x^{2}}{e^{x^{3}}}=x^{2}e^{-x^{3}}\).

Step2: Use substitution method

Let \(u = x^{3}\), then \(du=3x^{2}dx\), and \(x^{2}dx=\frac{1}{3}du\).
The integral \(\int x^{2}e^{-x^{3}}dx=\frac{1}{3}\int e^{-u}du\).

Step3: Integrate the new - form integral

According to the integral formula \(\int e^{ax}dx=\frac{1}{a}e^{ax}+C\) (\(a
eq0\)), for \(\int e^{-u}du\), we have \(\int e^{-u}du=-e^{-u}+C\).
Substituting back \(u = x^{3}\), we get \(\frac{1}{3}\int e^{-u}du=-\frac{1}{3}e^{-x^{3}}+C=-\frac{1}{3e^{x^{3}}}+C\).

Answer:

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