∫x^2+x+3/(x-1)^3 . . . do I do a partial fraction decomposition and how do I go about that?

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- Mar 22nd 2009, 09:37 AMswensonmIntegral
∫x^2+x+3/(x-1)^3 . . . do I do a partial fraction decomposition and how do I go about that?

- Mar 22nd 2009, 10:03 AMred_dog
$\displaystyle \frac{x^2+x+3}{(x-1)^3}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{(x-1)^3}$

- Mar 22nd 2009, 04:28 PMKrizalid
Put $\displaystyle u=x-1,$ written backwards as $\displaystyle x=u+1$ and $\displaystyle du=dx,$ thus the integral becomes $\displaystyle \int{\frac{(u+1)^{2}+u+1+3}{u^{3}}\,du},$ then do the algebra and turn that integral into many ones.