jmc

Let $S$ denote the given sum, so $$S=\frac{x^2}{x-1}+\frac{x^4}{x^2-1}+\cdots +\frac{x^{4020}}{x^{2010}-1}=\sum _{k=1}^{2010}\frac{x^{2k}}{x^k-1}.\text{\hspace{1em}(1)}$$We can reverse the order of the terms, to get $$S=\frac{x^{4020}}{x^{2010}-1}+\frac{x^{4018}}{x^{2009}-1}+\cdots +\frac{x^2}{x-1}=\sum _{k=1}^{2010}\frac{x^{4022-2k}}{x^{2011-k}-1}.$$Since $x^{2011}=1$, $$\frac{x^{4022-2k}}{x^{2011-k}-1}=\frac{x^{-2k}}{x^{-k}-1}=\frac{1}{x^k-x^{2k}}=\frac{1}{x^k(1-x^k)},$$so $$S=\sum _{k=1}^{2010}\frac{1}{x^k(1-x^k)}.\text{\hspace{1em}(2)}$$Adding equations (1) and (2), we get $$\begin{array}{rl}2S& =\sum _{k=1}^{2010}\frac{x^{2k}}{x^k-1}+\sum _{k=1}^{2010}\frac{1}{x^k(1-x^k)}\\ & =\sum _{k=1}^{2010}\left[\frac{x^{2k}}{x^k-1}+\frac{1}{x^k(1-x^k)}\right]\\ & =\sum _{k=1}^{2010}\left[\frac{x^{3k}}{x^k(x^k-1)}-\frac{1}{x^k(x^k-1)}\right]\\ & =\sum _{k=1}^{2010}\frac{x^{3k}-1}{x^k(x^k-1)}.\end{array}$$We can factor $x^{3k}-1$ as $(x^k-1)(x^{2k}+x^k+1)$, so $$\begin{array}{rl}2S& =\sum _{k=1}^{2010}\frac{(x^k-1)(x^{2k}+x^k+1)}{x^k(x^k-1)}\\ & =\sum _{k=1}^{2010}\frac{x^{2k}+x^k+1}{x^k}\\ & =\sum _{k=1}^{2010}\left(x^k+1+\frac{1}{x^k}\right)\\ & =\left(x+1+\frac{1}{x}\right)+\left(x^2+1+\frac{1}{x^2}\right)+\cdots +\left(x^{2010}+1+\frac{1}{x^{2010}}\right)\\ & =(x+x^2+\cdots +x^{2010})+2010+\frac{1}{x}+\frac{1}{x^2}+\cdots +\frac{1}{x^{2010}}.\end{array}$$Since $x^{2011}=1$, we have that $x^{2011}-1=0$, which factors as $$(x-1)(x^{2010}+x^{2009}+\cdots +x+1)=0.$$We know that $x\ne 1$, so we can divide both sides by $x-1$, to get $$x^{2010}+x^{2009}+\cdots +x+1=0.$$Then $$\begin{array}{rl}2S& =(x+x^2+\cdots +x^{2010})+2010+\frac{1}{x}+\frac{1}{x^2}+\cdots +\frac{1}{x^{2010}}\\ & =(x+x^2+\cdots +x^{2010})+2010+\frac{x^{2010}+x^{2009}+\cdots +x}{x^{2011}}\\ & =(-1)+2010+\frac{-1}{1}\\ & =2008,\end{array}$$so $S=\boxed{1004}$.