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

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

Let $$$u=\frac{x}{2} - 3$$$.

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

Therefore,

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

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=2$$$ and $$$f{\left(u \right)} = u^{5}$$$:

$${\color{red}{\int{2 u^{5} d u}}} = {\color{red}{\left(2 \int{u^{5} d u}\right)}}$$

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

$$2 {\color{red}{\int{u^{5} d u}}}=2 {\color{red}{\frac{u^{1 + 5}}{1 + 5}}}=2 {\color{red}{\left(\frac{u^{6}}{6}\right)}}$$

Recall that $$$u=\frac{x}{2} - 3$$$:

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

Therefore,

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

Simplify:

$$\int{\left(\frac{x}{2} - 3\right)^{5} d x} = \frac{\left(x - 6\right)^{6}}{192}$$

Add the constant of integration:

$$\int{\left(\frac{x}{2} - 3\right)^{5} d x} = \frac{\left(x - 6\right)^{6}}{192}+C$$

$$$\int \left(\frac{x}{2} - 3\right)^{5}\, dx = \frac{\left(x - 6\right)^{6}}{192} + C$$$A