$\int \frac{1}{x(\log x)^{2}} dx$ is equal to:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $-\frac{1}{\log x} + c$
Let $I = \int \frac{1}{x(\log x)^{2}} dx$
$\log x = t ⇒\frac{1}{x} dx = dt$
$I = \int \frac{dt}{t^{2}} = -\frac{1}{t} + c$
Thus, $I = -\frac{1}{\log x} + c$