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

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

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 $$$\frac{dx}{x} = du$$$.

The integral can be rewritten as

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

The integral of $$$\frac{1}{u}$$$ is $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

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

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

$$\ln{\left(\left|{{\color{red}{u}}}\right| \right)} = \ln{\left(\left|{{\color{red}{\ln{\left(x \right)}}}}\right| \right)}$$

Therefore,

$$\int{\frac{1}{x \ln{\left(x \right)}} d x} = \ln{\left(\left|{\ln{\left(x \right)}}\right| \right)}$$

Add the constant of integration:

$$\int{\frac{1}{x \ln{\left(x \right)}} d x} = \ln{\left(\left|{\ln{\left(x \right)}}\right| \right)}+C$$

$$$\int \frac{1}{x \ln\left(x\right)}\, dx = \ln\left(\left|{\ln\left(x\right)}\right|\right) + C$$$A