Integral of $$$\frac{z \left(6 - 2 z\right)^{5}}{3}$$$

Find $$$\int \frac{z \left(6 - 2 z\right)^{5}}{3}\, dz$$$.

Simplify the integrand:

$${\color{red}{\int{\frac{z \left(6 - 2 z\right)^{5}}{3} d z}}} = {\color{red}{\int{\frac{32 z \left(3 - z\right)^{5}}{3} d z}}}$$

Apply the constant multiple rule $$$\int c f{\left(z \right)}\, dz = c \int f{\left(z \right)}\, dz$$$ with $$$c=\frac{32}{3}$$$ and $$$f{\left(z \right)} = z \left(3 - z\right)^{5}$$$:

$${\color{red}{\int{\frac{32 z \left(3 - z\right)^{5}}{3} d z}}} = {\color{red}{\left(\frac{32 \int{z \left(3 - z\right)^{5} d z}}{3}\right)}}$$

Let $$$u=3 - z$$$.

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

Thus,

$$\frac{32 {\color{red}{\int{z \left(3 - z\right)^{5} d z}}}}{3} = \frac{32 {\color{red}{\int{u^{5} \left(u - 3\right) d u}}}}{3}$$

Expand the expression:

$$\frac{32 {\color{red}{\int{u^{5} \left(u - 3\right) d u}}}}{3} = \frac{32 {\color{red}{\int{\left(u^{6} - 3 u^{5}\right)d u}}}}{3}$$

Integrate term by term:

$$\frac{32 {\color{red}{\int{\left(u^{6} - 3 u^{5}\right)d u}}}}{3} = \frac{32 {\color{red}{\left(- \int{3 u^{5} d u} + \int{u^{6} d u}\right)}}}{3}$$

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=6$$$:

$$- \frac{32 \int{3 u^{5} d u}}{3} + \frac{32 {\color{red}{\int{u^{6} d u}}}}{3}=- \frac{32 \int{3 u^{5} d u}}{3} + \frac{32 {\color{red}{\frac{u^{1 + 6}}{1 + 6}}}}{3}=- \frac{32 \int{3 u^{5} d u}}{3} + \frac{32 {\color{red}{\left(\frac{u^{7}}{7}\right)}}}{3}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=3$$$ and $$$f{\left(u \right)} = u^{5}$$$:

$$\frac{32 u^{7}}{21} - \frac{32 {\color{red}{\int{3 u^{5} d u}}}}{3} = \frac{32 u^{7}}{21} - \frac{32 {\color{red}{\left(3 \int{u^{5} d u}\right)}}}{3}$$

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

$$\frac{32 u^{7}}{21} - 32 {\color{red}{\int{u^{5} d u}}}=\frac{32 u^{7}}{21} - 32 {\color{red}{\frac{u^{1 + 5}}{1 + 5}}}=\frac{32 u^{7}}{21} - 32 {\color{red}{\left(\frac{u^{6}}{6}\right)}}$$

Recall that $$$u=3 - z$$$:

$$- \frac{16 {\color{red}{u}}^{6}}{3} + \frac{32 {\color{red}{u}}^{7}}{21} = - \frac{16 {\color{red}{\left(3 - z\right)}}^{6}}{3} + \frac{32 {\color{red}{\left(3 - z\right)}}^{7}}{21}$$

Therefore,

$$\int{\frac{z \left(6 - 2 z\right)^{5}}{3} d z} = \frac{32 \left(3 - z\right)^{7}}{21} - \frac{16 \left(3 - z\right)^{6}}{3}$$

Simplify:

$$\int{\frac{z \left(6 - 2 z\right)^{5}}{3} d z} = \frac{16 \left(- 2 z - 1\right) \left(z - 3\right)^{6}}{21}$$

Add the constant of integration:

$$\int{\frac{z \left(6 - 2 z\right)^{5}}{3} d z} = \frac{16 \left(- 2 z - 1\right) \left(z - 3\right)^{6}}{21}+C$$

$$$\int \frac{z \left(6 - 2 z\right)^{5}}{3}\, dz = \frac{16 \left(- 2 z - 1\right) \left(z - 3\right)^{6}}{21} + C$$$A