Question

Choose the correct answer.
${\int }_1^{\sqrt{3}}\frac{1}{1+x^2}dx$ is equal to
$\left(a\right)\frac{\pi }{3}\left(b\right)\frac{2\pi }{3}\left(c\right)\frac{\pi }{6}\left(d\right)\frac{\pi }{12}$

Answer 1

${\int }_1^{\sqrt{3}}\frac{1}{1+x^2}dx={\int }_1^{\sqrt{3}}\frac{1}{x^2+1^2}dx={\left[\frac{1}{1}ta{n}^{-1}\right(\frac{x}{1}\left)\right]}_1^{\sqrt{3}}=ta{n}^{-1}\sqrt{3}-ta{n}^{-1}1(\because \int \frac{dx}{1+x^2}=ta{n}^{-1}x)=\frac{\pi }{3}-\frac{\pi }{4}\Rightarrow \frac{4\pi -3\pi }{12}=\frac{\pi }{12}.$ Hence, the option (d)is correct.