# Integral

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