Integral of $$$3 e x^{2} \left(2 x^{3} - 8\right)$$$

Find $$$\int 3 e x^{2} \left(2 x^{3} - 8\right)\, dx$$$.

Let $$$u=2 x^{3} - 8$$$.

Then $$$du=\left(2 x^{3} - 8\right)^{\prime }dx = 6 x^{2} dx$$$ (steps can be seen »), and we have that $$$x^{2} dx = \frac{du}{6}$$$.

Thus,

$${\color{red}{\int{3 e x^{2} \left(2 x^{3} - 8\right) d x}}} = {\color{red}{\int{\frac{e u}{2} d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{e}{2}$$$ and $$$f{\left(u \right)} = u$$$:

$${\color{red}{\int{\frac{e u}{2} d u}}} = {\color{red}{\left(\frac{e \int{u d u}}{2}\right)}}$$

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=1$$$:

$$\frac{e {\color{red}{\int{u d u}}}}{2}=\frac{e {\color{red}{\frac{u^{1 + 1}}{1 + 1}}}}{2}=\frac{e {\color{red}{\left(\frac{u^{2}}{2}\right)}}}{2}$$

Recall that $$$u=2 x^{3} - 8$$$:

$$\frac{e {\color{red}{u}}^{2}}{4} = \frac{e {\color{red}{\left(2 x^{3} - 8\right)}}^{2}}{4}$$

Therefore,

$$\int{3 e x^{2} \left(2 x^{3} - 8\right) d x} = \frac{e \left(2 x^{3} - 8\right)^{2}}{4}$$

Simplify:

$$\int{3 e x^{2} \left(2 x^{3} - 8\right) d x} = e \left(x^{3} - 4\right)^{2}$$

Add the constant of integration:

$$\int{3 e x^{2} \left(2 x^{3} - 8\right) d x} = e \left(x^{3} - 4\right)^{2}+C$$

$$$\int 3 e x^{2} \left(2 x^{3} - 8\right)\, dx = e \left(x^{3} - 4\right)^{2} + C$$$A