Question

The value of $\int \frac{{\tan }^{-1}\left(\ln x\right)+{\cot }^{-1}\left(\ln x\right)}{1+x^2}dx$ such that $x>1,$ is
(where $C$ of constant of integration)

• A. $\frac{\pi }{2}{\tan }^{-1}x+C$
• B. $\frac{\pi }{2}{\cot }^{-1}x+C$
• C. $\frac{\pi }{2}{\sec }^{-1}x+C$
• D. $\ln |1+x|+C$

Answer 1

The correct option is A $\frac{\pi }{2}{\tan }^{-1}x+C$

Let $I=\int \frac{{\tan }^{-1}\left(\ln x\right)+{\cot }^{-1}\left(\ln x\right)}{1+x^2}dx;x>1$
$I=\int \frac{\frac{\pi }{2}}{1+x^2}dx=\frac{\pi }{2}\int \frac{dx}{1+x^2}$

$\Rightarrow I=\frac{\pi }{2}{\tan }^{-1}x+C$