Integral of $$$x \left(2 x - 1\right)^{7}$$$

Find $$$\int x \left(2 x - 1\right)^{7}\, dx$$$.

Let $$$u=2 x - 1$$$.

Then $$$du=\left(2 x - 1\right)^{\prime }dx = 2 dx$$$ (steps can be seen »), and we have that $$$dx = \frac{du}{2}$$$.

The integral becomes

$${\color{red}{\int{x \left(2 x - 1\right)^{7} d x}}} = {\color{red}{\int{\frac{u^{7} \left(u + 1\right)}{4} d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{4}$$$ and $$$f{\left(u \right)} = u^{7} \left(u + 1\right)$$$:

$${\color{red}{\int{\frac{u^{7} \left(u + 1\right)}{4} d u}}} = {\color{red}{\left(\frac{\int{u^{7} \left(u + 1\right) d u}}{4}\right)}}$$

Expand the expression:

$$\frac{{\color{red}{\int{u^{7} \left(u + 1\right) d u}}}}{4} = \frac{{\color{red}{\int{\left(u^{8} + u^{7}\right)d u}}}}{4}$$

Integrate term by term:

$$\frac{{\color{red}{\int{\left(u^{8} + u^{7}\right)d u}}}}{4} = \frac{{\color{red}{\left(\int{u^{7} d u} + \int{u^{8} d u}\right)}}}{4}$$

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=7$$$:

$$\frac{\int{u^{8} d u}}{4} + \frac{{\color{red}{\int{u^{7} d u}}}}{4}=\frac{\int{u^{8} d u}}{4} + \frac{{\color{red}{\frac{u^{1 + 7}}{1 + 7}}}}{4}=\frac{\int{u^{8} d u}}{4} + \frac{{\color{red}{\left(\frac{u^{8}}{8}\right)}}}{4}$$

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=8$$$:

$$\frac{u^{8}}{32} + \frac{{\color{red}{\int{u^{8} d u}}}}{4}=\frac{u^{8}}{32} + \frac{{\color{red}{\frac{u^{1 + 8}}{1 + 8}}}}{4}=\frac{u^{8}}{32} + \frac{{\color{red}{\left(\frac{u^{9}}{9}\right)}}}{4}$$

Recall that $$$u=2 x - 1$$$:

$$\frac{{\color{red}{u}}^{8}}{32} + \frac{{\color{red}{u}}^{9}}{36} = \frac{{\color{red}{\left(2 x - 1\right)}}^{8}}{32} + \frac{{\color{red}{\left(2 x - 1\right)}}^{9}}{36}$$

Therefore,

$$\int{x \left(2 x - 1\right)^{7} d x} = \frac{\left(2 x - 1\right)^{9}}{36} + \frac{\left(2 x - 1\right)^{8}}{32}$$

Simplify:

$$\int{x \left(2 x - 1\right)^{7} d x} = \frac{\left(2 x - 1\right)^{8} \left(16 x + 1\right)}{288}$$

Add the constant of integration:

$$\int{x \left(2 x - 1\right)^{7} d x} = \frac{\left(2 x - 1\right)^{8} \left(16 x + 1\right)}{288}+C$$

$$$\int x \left(2 x - 1\right)^{7}\, dx = \frac{\left(2 x - 1\right)^{8} \left(16 x + 1\right)}{288} + C$$$A