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What is the Integral of # tan^-1x/(1+x^2)#?

1 Answer

Aug 3, 2016

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

Explanation:

#int \ tan^-1x/(1+x^2) \ dx#

#= int \ tan^-1 x * d/dx (tan^(-1) x) \ dx#

#= int d/dx (1/2(\ tan^-1 x )^2 ) \ dx#

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

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