Integral of $$$x^{2} \left(6 - x^{3}\right)^{5}$$$

Find $$$\int x^{2} \left(6 - x^{3}\right)^{5}\, dx$$$.

Let $$$u=6 - x^{3}$$$.

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

So,

$${\color{red}{\int{x^{2} \left(6 - x^{3}\right)^{5} d x}}} = {\color{red}{\int{\left(- \frac{u^{5}}{3}\right)d u}}}$$

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 $$$f{\left(u \right)} = u^{5}$$$:

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

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=5$$$:

$$- \frac{{\color{red}{\int{u^{5} d u}}}}{3}=- \frac{{\color{red}{\frac{u^{1 + 5}}{1 + 5}}}}{3}=- \frac{{\color{red}{\left(\frac{u^{6}}{6}\right)}}}{3}$$

Recall that $$$u=6 - x^{3}$$$:

$$- \frac{{\color{red}{u}}^{6}}{18} = - \frac{{\color{red}{\left(6 - x^{3}\right)}}^{6}}{18}$$

Therefore,

$$\int{x^{2} \left(6 - x^{3}\right)^{5} d x} = - \frac{\left(6 - x^{3}\right)^{6}}{18}$$

Simplify:

$$\int{x^{2} \left(6 - x^{3}\right)^{5} d x} = - \frac{\left(x^{3} - 6\right)^{6}}{18}$$

Add the constant of integration:

$$\int{x^{2} \left(6 - x^{3}\right)^{5} d x} = - \frac{\left(x^{3} - 6\right)^{6}}{18}+C$$

$$$\int x^{2} \left(6 - x^{3}\right)^{5}\, dx = - \frac{\left(x^{3} - 6\right)^{6}}{18} + C$$$A