Question

The value of the definite integral $\frac{1}{\pi }\underset{0}{\overset{2}{\int }}\frac{{\tan }^{-1}\left(\frac{x}{2}\right)}{x^2+4}dx$ is equal to

• A. $\frac{{\pi }^2}{64}$
• B. $\frac{{\pi }^2}{32}$
• C. $\frac{\pi }{64}$
• D. $\frac{\pi }{32}$

Answer 1

Answer: (C)

Put ${\tan }^{-1}\left(\frac{x}{2}\right)=t$ an proceed