Question

$\int^{\sqrt{3}}_{0} {\frac{dx}{1+x^2}$
(A)\frac{\pi}{3}[/tex]
(B)\frac{2\pi}{3}[/tex]
(C)\frac{\pi}{6}[/tex]
(D)\frac{\pi}{12}[/tex]

Answer 1

Answer:

option (A) is correct

Step-by-step explanation:

$Concept:\\\\\int{\frac{1}{1+x^2}}\,dx=tan^{-1}x+c\\\\tan\frac{\pi}{3}=\sqrt{3}\\tan0=0$

Now,

$\int{\limits^{\sqrt{3}}_0{\frac{1}{1+x^2}}\:dx\\\\=[tan^{-1}x]^{\sqrt{3}}_0\\\\=tan^{-1}\sqrt{3}-tan^{-1}0\\\\=\frac{\pi}{3}-0\\\\=\frac{\pi}{3}$