Integral of $$$\frac{\ln^{5}\left(x\right)}{x}$$$

Find $$$\int \frac{\ln^{5}\left(x\right)}{x}\, dx$$$.

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

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

Thus,

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

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

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

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

$$\frac{{\color{red}{u}}^{6}}{6} = \frac{{\color{red}{\ln{\left(x \right)}}}^{6}}{6}$$

Therefore,

$$\int{\frac{\ln{\left(x \right)}^{5}}{x} d x} = \frac{\ln{\left(x \right)}^{6}}{6}$$

Add the constant of integration:

$$\int{\frac{\ln{\left(x \right)}^{5}}{x} d x} = \frac{\ln{\left(x \right)}^{6}}{6}+C$$

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