How do you integrate #tan^-1x/(1+x^2)#?

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Answer 1 · 2016-12-24T21:26:09

# tan^-1x/2 + C#

Let #u = tan^-1x#. Then #du = 1/(x^2 + 1)dx# and #dx = (du)/(1/(x^2 + 1)) = x^2 + 1du#

Substitute:

#=> intu/(1 + x^2) * 1 + x^2 du#

#=> intu du#

Use the power rule #int(x^n) = x^(n + 1)/(n + 1) + C#.

#=> 1/2u^2 + C#

#=> 1/2(tan^-1x) + C#

#=> tan^-1x/2 + C#

Hopefully this helps!