Integral of $$$\frac{1}{x \ln^{3}\left(x\right)}$$$

Find $$$\int \frac{1}{x \ln^{3}\left(x\right)}\, dx$$$.

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,

$${\color{red}{\int{\frac{1}{x \ln{\left(x \right)}^{3}} d x}}} = {\color{red}{\int{\frac{1}{u^{3}} d u}}}$$

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=-3$$$:

$${\color{red}{\int{\frac{1}{u^{3}} d u}}}={\color{red}{\int{u^{-3} d u}}}={\color{red}{\frac{u^{-3 + 1}}{-3 + 1}}}={\color{red}{\left(- \frac{u^{-2}}{2}\right)}}={\color{red}{\left(- \frac{1}{2 u^{2}}\right)}}$$

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

$$- \frac{{\color{red}{u}}^{-2}}{2} = - \frac{{\color{red}{\ln{\left(x \right)}}}^{-2}}{2}$$

Therefore,

$$\int{\frac{1}{x \ln{\left(x \right)}^{3}} d x} = - \frac{1}{2 \ln{\left(x \right)}^{2}}$$

Add the constant of integration:

$$\int{\frac{1}{x \ln{\left(x \right)}^{3}} d x} = - \frac{1}{2 \ln{\left(x \right)}^{2}}+C$$

$$$\int \frac{1}{x \ln^{3}\left(x\right)}\, dx = - \frac{1}{2 \ln^{2}\left(x\right)} + C$$$A