How to integrate $\int \frac{x-1}{x^2\ln(x)+x}dx$

Question

According to Wolfram Alpha, $$\int \frac{x-1}{x^2\ln(x)+x}dx = \ln\left(\ln(x)+\frac{1}{x}\right)$$ but I don't really know how to do this.

Answer 1

We have $$ \frac{x-1}{x^2\log x+x}=\frac{\frac1x-\frac1{x^2}}{\log x+\frac1x} $$ is of the form $\dfrac{f'(x)}{f(x)}$, so integrating with respect to $x$ gives the answer.

Answer 2

It requires dividing the top and bottom of your fraction by $x^2$. The resulting numerator will be the derivative of the resulting denominator, so the whole will be $\int du/u$ where $u$ is the denominator, hence $\ln u$.