For \(x \gt 1\), \(\int\displaystyle\frac{dx}{x(\log x)^2}\) is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle -\frac{1}{\log x}+C\)
The factor \(\frac{1}{x}\) is the derivative of \(\log x\), so substitute
\( \displaystyle t = \log x,\quad dt = \frac{dx}{x} \)
Then
\( \displaystyle \int\frac{dx}{x(\log x)^2} = \int t^{-2}\,dt \)
\( \displaystyle = -t^{-1}+C = -\frac{1}{\log x}+C \)
Why the other options are incorrect
Option (2) \(\displaystyle \frac{1}{\log x}+C\) — Incorrect. The sign slip: \(\int t^{-2}\,dt = \frac{t^{-1}}{-1} = -\frac{1}{t}\).
Option (3) \(\displaystyle \frac{(\log x)^3}{3}+C\) — Incorrect. That is \(\int t^2\,dt\); here \(\log x\) is squared in the denominator, so the power is \(-2\).
Option (4) \(2\log|\log x|+C\) — Incorrect. The log rule applies only to \(t^{-1}\), not to \(t^{-2}\).