Question
Evaluate the definite integral in Exercise:

$\int\limits_{0}^{1}\frac{\text{dx}}{1+\text{x}^{2}}$

Answer

$\text{Let}\ \text{I}=\int\limits_{0}^{4}\frac{\text{dx}}{1+\text{x}^{2}}$
$\int\frac{\text{dx}}{1+\text{x}^{2}}=\tan^{-1}\text{x}=\text{F}\text{(x)}$

By second fundamental theorem of calculus, we obtain

$\text{I}=\text{F}(1)-\text{F}(0)$ 

$=\tan^{-1}(1)-\tan^{-1}(0)$

$=\frac{\pi}{4}$

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