Integral of $$$\left(d + e x\right)^{7}$$$ with respect to $$$x$$$

Find $$$\int \left(d + e x\right)^{7}\, dx$$$.

Let $$$u=d + e x$$$.

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

So,

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

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

$${\color{red}{\int{\frac{u^{7}}{e} d u}}} = {\color{red}{\frac{\int{u^{7} d u}}{e}}}$$

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

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

Recall that $$$u=d + e x$$$:

$$\frac{{\color{red}{u}}^{8}}{8 e} = \frac{{\color{red}{\left(d + e x\right)}}^{8}}{8 e}$$

Therefore,

$$\int{\left(d + e x\right)^{7} d x} = \frac{\left(d + e x\right)^{8}}{8 e}$$

Add the constant of integration:

$$\int{\left(d + e x\right)^{7} d x} = \frac{\left(d + e x\right)^{8}}{8 e}+C$$

$$$\int \left(d + e x\right)^{7}\, dx = \frac{\left(d + e x\right)^{8}}{8 e} + C$$$A