What is the antiderivative of $\frac{1}{1 + x^{2}}$ $1/(1+x^2)$ ?

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Answer 1 · 2018-03-24T22:56:25

${tan}^{- 1} (x) + C$ $tan^-1(x)+C$

We want a function which, when differentiated, will yield $\frac{1}{1 + x^{2}} .$ $1/(1+x^2).$

${tan}^{- 1} (x)$ $tan^-1(x)$ fulfills this requirement, as $\frac{d}{d x} {tan}^{- 1} (x) = \frac{1}{1 + x^{2}} .$ $d/dxtan^-1(x)=1/(1+x^2).$

We also add in an arbitrary constant $C,$ $C,$ any constant could be part of our antiderivative, as differentiating a constant yields zero.

What is the antiderivative of 11+x21/(1+x^2)?