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

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

The input is rewritten: $$$\int{\frac{1}{x \ln{\left(x^{3} \right)}} d x}=\int{\frac{1}{3 x \ln{\left(x \right)}} d x}$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{3}$$$ and $$$f{\left(x \right)} = \frac{1}{x \ln{\left(x \right)}}$$$:

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

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 becomes

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

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

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

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

$$\frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{3} = \frac{\ln{\left(\left|{{\color{red}{\ln{\left(x \right)}}}}\right| \right)}}{3}$$

Therefore,

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

Add the constant of integration:

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

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