Question

If ${\int }_0^b\frac{dx}{1+x^2}={\int }_b^{\infty }\frac{dx}{1+x^2}$, then $b=$

• A. ${\tan }^{-1}\left(\frac{1}{3}\right)$
• B. $\frac{\sqrt{3}}{2}$
• C. $\sqrt{3}$
• D. $1$

Answer 1

The correct option is D $1$

${\int }_0^b\frac{dx}{1+x^2}={\int }_b^{\infty }\frac{dx}{1+x^2}$

$\Rightarrow [{\tan }^{-1}x{]}_0^b=[{\tan }^{-1}x{]}_b^{\infty }$

$\Rightarrow {\tan }^{-1}b-0=\frac{\pi }{2}-{\tan }^{-1}b$

$\Rightarrow {\tan }^{-1}b=\frac{\pi }{4}$

$\Rightarrow b=\tan \frac{\pi }{4}=1$