What is the integral of #arctan(x)#?

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Answer 1 · 2018-08-12T16:30:29

#inttan^(-1)(x)dx=xtan^(-1)(x)-1/2ln(1+x^2)+C#, #C in RR#

#I=inttan^(-1)(x)dx#

#f(x)=tan^(-1)(x)#, #f'(x)=1/(1+x^2)#

#g'(x)=1#, #g(x)=x#

#I=xtan^(-1)(x)-intx/(1+x^2)dx#

#=xtan^(-1)(x)-1/2int(2x)/(1+x^2)dx#

Let #u=1+x^2#

#du=2xdx#

#I=xtan^(-1)(x)-1/2int1/udu#

#=xtan^(-1)(x)-1/2ln(|u|)#

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

\0/ Here's our answer !