Question

Integrate:
${\int }_1^{\sqrt{3}}\frac{dx}{1+x^2}$ equals

Answer 1

${\int }_1^{\sqrt{3}}\frac{dx}{1+x^2}$

$\Rightarrow {{\left[{\mathrm{Tan}}^{-1}x\right]}_1}^{\sqrt{3}}$

$\Rightarrow \left[{\mathrm{Tan}}^{-1}\left[\sqrt{3}\right]-{\mathrm{Tan}}^{-1}\left[1\right]\right]$

$\Rightarrow \left[\frac{\pi }{3}-\frac{\pi }{4}\right]$

$=\left[\frac{4\pi -3\pi }{12}\right]$

$=\frac{\pi }{12}Answer$