What is the antiderivative of $p(x) = 1/ (x^2 + 1)$ ?

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Answer 1 · 2016-06-27T08:55:29

$= arctan x + C$

$int \ 1/ (x^2 + 1) \ dx$

this is very well known and widely used integral

use the sub $x = tan t, \qquad dx = sec^2 t \ dt$

so we have

$int \ 1/ (tan^2 t + 1) \ sec^2 t \ dt$

$= int \ (sec^2 t)/(sec^2 t) \ dt = int \ dt$

$= t + C$

$= arctan x + C$

it is built on the well-known trig identity $tan^2 varphi + 1 = sec^2 varphi$

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