Question

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

Answer 1

Let $I={\int }_0^1\frac{dx}{1+x^2}$
$\Rightarrow \int \frac{dx}{1+x^2}={\tan }^{-1}x=F\left(x\right)$
By second fundamental theorem of calculus, we obtain
$I=F\left(1\right)-F\left(0\right)$
$={\tan }^{-1}\left(1\right)-{\tan }^{-1}\left(0\right)=\frac{\pi }{4}$