# What is the antiderivative of 1/(1+x^2)?

${\tan}^{-} 1 \left(x\right) + C$
We want a function which, when differentiated, will yield $\frac{1}{1 + {x}^{2}} .$
${\tan}^{-} 1 \left(x\right)$ fulfills this requirement, as $\frac{d}{\mathrm{dx}} {\tan}^{-} 1 \left(x\right) = \frac{1}{1 + {x}^{2}} .$
We also add in an arbitrary constant $C ,$ any constant could be part of our antiderivative, as differentiating a constant yields zero.