CUET Preparation Today
$\int \frac{1}{x(\log x)^{2}} dx$ is equal to: |
$2 \log (\log x) + c$ $-\frac{1}{\log x} + c$ $\frac{(\log x)^{3}}{3} + c$ $-\frac{3}{(\log)^{3}} + c$ |
$-\frac{1}{\log x} + c$ |
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, $I = -\frac{1}{\log x} + c$ |
