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

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

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