Integral of $$$\left(3 x - 4\right)^{5}$$$

Find $$$\int \left(3 x - 4\right)^{5}\, dx$$$.

Let $$$u=3 x - 4$$$.

Then $$$du=\left(3 x - 4\right)^{\prime }dx = 3 dx$$$ (steps can be seen »), and we have that $$$dx = \frac{du}{3}$$$.

Thus,

$${\color{red}{\int{\left(3 x - 4\right)^{5} d x}}} = {\color{red}{\int{\frac{u^{5}}{3} d u}}}$$

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

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

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 $$$u=3 x - 4$$$:

$$\frac{{\color{red}{u}}^{6}}{18} = \frac{{\color{red}{\left(3 x - 4\right)}}^{6}}{18}$$

Therefore,

$$\int{\left(3 x - 4\right)^{5} d x} = \frac{\left(3 x - 4\right)^{6}}{18}$$

Add the constant of integration:

$$\int{\left(3 x - 4\right)^{5} d x} = \frac{\left(3 x - 4\right)^{6}}{18}+C$$

$$$\int \left(3 x - 4\right)^{5}\, dx = \frac{\left(3 x - 4\right)^{6}}{18} + C$$$A