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 #