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

1 Answer

Mar 9, 2018

The answer is #=(arctanx)^2/2+C#

Explanation:

Apply the substitution

#u=arctanx#, #=>#, #du=(dx)/(1+x^2)#

Therefore,

#int(arctanxdx)/(1+x^2)=int(u*(1+x^2)du)/(1+x^2)=intudu#

#=u^2/2#

#=(arctanx)^2/2+C#

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