How do you find the integral of #ln(1+x^2)#?

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Answer 1 · 2018-06-03T05:58:01

The answer is #=xln(1+x^2)-2x+2arctanx+C#

Calculate this integral by integration by parts

#intuv'=uv-intu'v#

#u=ln(1+x^2)#, #=>#, #u'=(2x)/(1+x^2)#

#v'=1#, #=>#, #v=x#

Therefore, the integral is

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

The second integral is

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

#=2int1dx-2int(1dx)/(1+x^2)#

#=2x-2arctanx#

Finally, the integral is

#I=xln(1+x^2)-2x+2arctanx+C#