Integral of $$$x^{2} e^{x^{3} - 5}$$$

Find $$$\int x^{2} e^{x^{3} - 5}\, dx$$$.

Let $$$u=x^{3} - 5$$$.

Then $$$du=\left(x^{3} - 5\right)^{\prime }dx = 3 x^{2} dx$$$ (steps can be seen »), and we have that $$$x^{2} dx = \frac{du}{3}$$$.

So,

$${\color{red}{\int{x^{2} e^{x^{3} - 5} d x}}} = {\color{red}{\int{\frac{e^{u}}{3} d u}}}$$

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 $$$u=x^{3} - 5$$$:

$$\frac{e^{{\color{red}{u}}}}{3} = \frac{e^{{\color{red}{\left(x^{3} - 5\right)}}}}{3}$$

Therefore,

$$\int{x^{2} e^{x^{3} - 5} d x} = \frac{e^{x^{3} - 5}}{3}$$

Add the constant of integration:

$$\int{x^{2} e^{x^{3} - 5} d x} = \frac{e^{x^{3} - 5}}{3}+C$$

$$$\int x^{2} e^{x^{3} - 5}\, dx = \frac{e^{x^{3} - 5}}{3} + C$$$A