Integral of $$$x^{2} \left(2 x^{3} + 3\right)^{3}$$$

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

Let $$$u=2 x^{3} + 3$$$.

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

The integral becomes

$${\color{red}{\int{x^{2} \left(2 x^{3} + 3\right)^{3} d x}}} = {\color{red}{\int{\frac{u^{3}}{6} d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{6}$$$ and $$$f{\left(u \right)} = u^{3}$$$:

$${\color{red}{\int{\frac{u^{3}}{6} d u}}} = {\color{red}{\left(\frac{\int{u^{3} d u}}{6}\right)}}$$

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=3$$$:

$$\frac{{\color{red}{\int{u^{3} d u}}}}{6}=\frac{{\color{red}{\frac{u^{1 + 3}}{1 + 3}}}}{6}=\frac{{\color{red}{\left(\frac{u^{4}}{4}\right)}}}{6}$$

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

$$\frac{{\color{red}{u}}^{4}}{24} = \frac{{\color{red}{\left(2 x^{3} + 3\right)}}^{4}}{24}$$

Therefore,

$$\int{x^{2} \left(2 x^{3} + 3\right)^{3} d x} = \frac{\left(2 x^{3} + 3\right)^{4}}{24}$$

Add the constant of integration:

$$\int{x^{2} \left(2 x^{3} + 3\right)^{3} d x} = \frac{\left(2 x^{3} + 3\right)^{4}}{24}+C$$

$$$\int x^{2} \left(2 x^{3} + 3\right)^{3}\, dx = \frac{\left(2 x^{3} + 3\right)^{4}}{24} + C$$$A