# Integral

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• March 22nd 2009, 09:37 AM
swensonm
Integral
x^2+x+3/(x-1)^3 . . . do I do a partial fraction decomposition and how do I go about that?
• March 22nd 2009, 10:03 AM
red_dog
$\frac{x^2+x+3}{(x-1)^3}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{(x-1)^3}$
• March 22nd 2009, 04:28 PM
Krizalid
Put $u=x-1,$ written backwards as $x=u+1$ and $du=dx,$ thus the integral becomes $\int{\frac{(u+1)^{2}+u+1+3}{u^{3}}\,du},$ then do the algebra and turn that integral into many ones.