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how to do you solve ∫ [(x-1)/(x^2+1)]. take note that (x^2)+1
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it's u du substitution
first separate the integrals....do integral of x/x^2+1 - integral of 1/x^2+1
now use u du on the first integral letting u be x^2 + 1....take the derivative to get 1/2du=xdx
so you get 1/2 ln(u) and sub back in the x^2+1 for u
just by looking you can see the 2nd integral is tan^-1(x)
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so you can piece it together and get 1/2ln(x^2+1)-tan^-1(x)
i think...someone really should verify this...i'm still waking up
lol, thanks, u got it right. just im a lil bit confused on how you get the second integral to be tan^-1(x).
it's just observation....the derivative of tan-1 is 1/x^2+1 from back in calc 1....so you should be able to just see the integral and do the opposite
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