Integral of $$$\frac{1}{x^{2} \ln\left(x\right)}$$$

Find $$$\int \frac{1}{x^{2} \ln\left(x\right)}\, dx$$$.

Let $$$u=\frac{1}{x}$$$.

Then $$$du=\left(\frac{1}{x}\right)^{\prime }dx = - \frac{1}{x^{2}} dx$$$ (steps can be seen »), and we have that $$$\frac{dx}{x^{2}} = - du$$$.

Therefore,

$${\color{red}{\int{\frac{1}{x^{2} \ln{\left(x \right)}} d x}}} = {\color{red}{\int{\frac{1}{\ln{\left(u \right)}} d u}}}$$

This integral (Logarithmic Integral) does not have a closed form:

$${\color{red}{\int{\frac{1}{\ln{\left(u \right)}} d u}}} = {\color{red}{\operatorname{li}{\left(u \right)}}}$$

Recall that $$$u=\frac{1}{x}$$$:

$$\operatorname{li}{\left({\color{red}{u}} \right)} = \operatorname{li}{\left({\color{red}{\frac{1}{x}}} \right)}$$

Therefore,

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

Add the constant of integration:

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

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