Question

The value of $\underset{x\to \infty }{lim}\frac{{\int }_0^x{\left({\tan }^{-1}x\right)}^{2}dx}{\sqrt{x^2+1}}$ is?

• A. $\pi /4$
• B. ${\pi }^2/2$
• C. ${\pi }^2/4$
• D. $\pi$

Answer 1

The correct option is C ${\pi }^2/4$

$\underset{x\to \infty }{lim}\frac{{\int }_0^x({\text{tan}}^{-1}x{)}^{2}dx}{\sqrt{x^2+1}}$

Apply L hospital Rule

$=\underset{x\to \infty }{lim}\frac{({\text{tan}}^{-1}x{)}^2}{(\frac{x}{\sqrt{x^2+1}})}=\frac{{\pi }^2}{4}$