∫(x-1)/(x(x-log x))dx equals :( consider: log x=log_e x )

The correct answer is Option (1) → \( \log|x-\log x|+C \), \( C \) is an arbitrary constant.

The numerator is the derivative of the bracket in the denominator, once \( \frac{1}{x} \) is taken with it:

    \( \displaystyle \frac{d}{dx}(x-\log x)=1-\frac{1}{x}=\frac{x-1}{x} \)

So let \( t=x-\log x \), \( dt=\frac{x-1}{x}\,dx \):

    \( \displaystyle \int\frac{x-1}{x(x-\log x)}\,dx \)

    \( \displaystyle =\int\frac{dt}{t} \)

    \( \displaystyle =\log|t|+C \)

    \( \displaystyle =\log|x-\log x|+C \)

This is the form \( \int\frac{f'(x)}{f(x)}\,dx=\log|f(x)|+C \).