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Q. The value of $\int \frac{\sec^2 (\log \, x)}{x} dx $ is :

Integrals

Solution:

Let $I = \int \frac{\sec^2 (\log \, x)}{x} dx$
Put $\log \, x = t \Rightarrow \frac{1}{x} dx = dt$
$\therefore \, \int \sec^2 t \, dt = \tan \, t + c = \tan (\log \, x) + c $