Integral of $$$x^{2} e^{x^{3}}$$$
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Find $$$\int x^{2} e^{x^{3}}\, dx$$$.
Solution
Let $$$u=x^{3}$$$.
Then $$$du=\left(x^{3}\right)^{\prime }dx = 3 x^{2} dx$$$ (steps can be seen »), and we have that $$$x^{2} dx = \frac{du}{3}$$$.
Thus,
$${\color{red}{\int{x^{2} e^{x^{3}} d x}}} = {\color{red}{\int{\frac{e^{u}}{3} d u}}}$$
Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{3}$$$ and $$$f{\left(u \right)} = e^{u}$$$:
$${\color{red}{\int{\frac{e^{u}}{3} d u}}} = {\color{red}{\left(\frac{\int{e^{u} d u}}{3}\right)}}$$
The integral of the exponential function is $$$\int{e^{u} d u} = e^{u}$$$:
$$\frac{{\color{red}{\int{e^{u} d u}}}}{3} = \frac{{\color{red}{e^{u}}}}{3}$$
Recall that $$$u=x^{3}$$$:
$$\frac{e^{{\color{red}{u}}}}{3} = \frac{e^{{\color{red}{x^{3}}}}}{3}$$
Therefore,
$$\int{x^{2} e^{x^{3}} d x} = \frac{e^{x^{3}}}{3}$$
Add the constant of integration:
$$\int{x^{2} e^{x^{3}} d x} = \frac{e^{x^{3}}}{3}+C$$
Answer
$$$\int x^{2} e^{x^{3}}\, dx = \frac{e^{x^{3}}}{3} + C$$$A