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

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Answer 1 · 2018-04-28T14:35:41

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

We need

#int(u'(x)dx)/(u(x))=ln(u(x))+C#

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

The integral is

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

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

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