What is $int ( 1+x)/(1+x^2)dx$ ?

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Answer 1 · 2018-03-23T06:49:06

The answer is $=arctan(x)+1/2ln(1+x^2)+C$

The integral is

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

$I=I_1+I_2$

$I_1$ is a standard integral

$I_1=arctan(x)$

For $I_2$ , perform the substitution

$u=1+x^2$ , $=>$ , $du=2xdx$

Therefore,

$I_2=1/2int(du)/(u)=1/2ln(u)$

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

And finally,

$int((1+x)dx)/(1+x^2)=arctan(x)+1/2ln(1+x^2)+C$

What is int ( 1+x)/(1+x^2)dxint ( 1+x)/(1+x^2)dx?