$\int x^{2}e^{x^{3}}\,dx$ equals
- $\dfrac{1}{3}e^{x^{2}}+C$
- $\dfrac{1}{2}e^{x^{3}}+C$
- $\dfrac{1}{3}e^{x^{3}}+C$
- $\dfrac{1}{2}e^{x^{2}}+C$
Show answer and solution
Correct answer: C. $\dfrac{1}{3}e^{x^{3}}+C$
1. Put $t=x^{3}$, so $dt=3x^{2}\,dx$, i.e. $x^{2}\,dx=\dfrac{1}{3}dt$.
2. Integral $=\dfrac{1}{3}\int e^{t}\,dt=\dfrac{1}{3}e^{t}$.
3. Back-substitute: $\dfrac{1}{3}e^{x^{3}}+C$.
4. Options with $e^{x^{2}}$ misread the exponent; option B has wrong constant.
_Source: syllabus Mathematics Ch 7 "Integrals", p.329_