Question

${\int }_0^{60}\left[\frac{3}{x^2+1}\right]dx$, where [.] denotes the greatest integer function is equal to

• A. $3{\tan }^{-1}\sqrt{2}$
• B. $\sqrt{2}+1$
• C. $\frac{3}{\sqrt{2}}$
• D. infinite

Answer 1

The correct option is A $3{\tan }^{-1}\sqrt{2}$

${\int }_0^{60}\left[\frac{3}{x^2+1}\right]dx$

$={\int }_0^{\sqrt{2}}\left[\frac{3}{x^2+1}\right]dx+{\int }_{\sqrt{2}}^{60}\left[\right[\frac{3}{x^2+1}]dx$

$={\int }_0^{\sqrt{2}}\left[\frac{3}{x^2+1}\right]dx+0$

$=3{\int }_0^{\sqrt{2}}\frac{1}{x^2+1}dx$

$=3\left({\tan }^{-1}{]}_0^{\sqrt{2}}\right)$

$=3({\tan }^{-1}\sqrt{2}-0)=3{\tan }^{-1}\sqrt{2}$