∫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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$\displaystyle \frac{x^2+x+3}{(x-1)^3}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{(x-1)^3}$
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.
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