Question
Evaluate the following integrals:$\int\limits^1_0\frac{1}{1+\text{x}^2}\text{ dx}$

Answer

$\int\limits^1_0\frac{1}{1+\text{x}^2}\text{ dx}$$=\big[\tan^{-1}\text{x}\big]^1_0$
$=\tan^{-1}1-\tan^{-1}0$
$=\frac{\pi}{4}-0$
$=\frac{\pi}{4}$

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