For \(x \gt 0\), \(\int\displaystyle\frac{x - 1}{x(x - \log x)}\,dx\) is equal to (where C is a constant of integration)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\log|x - \log x| + C\)
Dividing the numerator and the denominator by x,
\( \displaystyle \frac{x - 1}{x(x - \log x)} = \frac{1 - \frac{1}{x}}{x - \log x} \)
Let \(t = x - \log x\). Then
\( \displaystyle dt = \left(1 - \frac{1}{x}\right)dx \)
So the numerator is the derivative of the denominator:
\( \displaystyle \int \frac{dt}{t} = \log|t| + C \)
\( = \log|x - \log x| + C \)