Integral of $$$\frac{\ln^{7}\left(z\right)}{z}$$$

Find $$$\int \frac{\ln^{7}\left(z\right)}{z}\, dz$$$.

Let $$$u=\ln{\left(z \right)}$$$.

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

The integral can be rewritten as

$${\color{red}{\int{\frac{\ln{\left(z \right)}^{7}}{z} d z}}} = {\color{red}{\int{u^{7} d u}}}$$

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

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

Recall that $$$u=\ln{\left(z \right)}$$$:

$$\frac{{\color{red}{u}}^{8}}{8} = \frac{{\color{red}{\ln{\left(z \right)}}}^{8}}{8}$$

Therefore,

$$\int{\frac{\ln{\left(z \right)}^{7}}{z} d z} = \frac{\ln{\left(z \right)}^{8}}{8}$$

Add the constant of integration:

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

$$$\int \frac{\ln^{7}\left(z\right)}{z}\, dz = \frac{\ln^{8}\left(z\right)}{8} + C$$$A