How do you evaluate: indefinite integral #(1+x)/(1+x^2) dx#?

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Answer 1 · 2018-03-24T09:38:11

The answer is #=arctanx+1/2ln(1+x^2)+C#

The integral is

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

The first integral

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

And the second integral is

#int(xdx)/(1+x^2)=1/2int(2xdx)/(1+x^2)#

#=1/2ln(1+x^2)#

And finally

#I=arctanx+1/2ln(1+x^2)+C#