Question

Evaluate the definite integrals.
${\int }_0^1\frac{1}{1+x^2}dx.$

Answer 1

${\int }_0^1\frac{1}{1+x^2}dx=[ta{n}^{-1}x{]}_0^1(\because \int \frac{d}{1+x^2}=ta{n}^{-1}x)=ta{n}^{-1}1-ta{n}^{-1}0=\frac{\pi }{4}$