Integral of $$$x^{5} \left(x^{6} - 7\right)^{4}$$$

Find $$$\int x^{5} \left(x^{6} - 7\right)^{4}\, dx$$$.

Let $$$u=x^{6} - 7$$$.

Then $$$du=\left(x^{6} - 7\right)^{\prime }dx = 6 x^{5} dx$$$ (steps can be seen »), and we have that $$$x^{5} dx = \frac{du}{6}$$$.

The integral becomes

$${\color{red}{\int{x^{5} \left(x^{6} - 7\right)^{4} d x}}} = {\color{red}{\int{\frac{u^{4}}{6} d u}}}$$

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 $$$f{\left(u \right)} = u^{4}$$$:

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

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=4$$$:

$$\frac{{\color{red}{\int{u^{4} d u}}}}{6}=\frac{{\color{red}{\frac{u^{1 + 4}}{1 + 4}}}}{6}=\frac{{\color{red}{\left(\frac{u^{5}}{5}\right)}}}{6}$$

Recall that $$$u=x^{6} - 7$$$:

$$\frac{{\color{red}{u}}^{5}}{30} = \frac{{\color{red}{\left(x^{6} - 7\right)}}^{5}}{30}$$

Therefore,

$$\int{x^{5} \left(x^{6} - 7\right)^{4} d x} = \frac{\left(x^{6} - 7\right)^{5}}{30}$$

Add the constant of integration:

$$\int{x^{5} \left(x^{6} - 7\right)^{4} d x} = \frac{\left(x^{6} - 7\right)^{5}}{30}+C$$

$$$\int x^{5} \left(x^{6} - 7\right)^{4}\, dx = \frac{\left(x^{6} - 7\right)^{5}}{30} + C$$$A