$\frac{P(x)}{Q(x)}-\frac{P(x)+1}{Q(x+1)}=\frac 1{x(x+2)}$$\Longrightarrow x(x+2)(P(x)Q(x+1)-P(x+1)Q(x))=Q(x)Q(x+1)$ for infinite many $x\in R$ , so the equality holds for all $x\in R$, because of $P,Q$ are polynomials. I denote $F(x)=\frac{P(x)}{Q(x)}$, so $F(x+1)=F(x) - \frac12.(\frac 1x-\frac 1{x+2})$,..., $F(x+n+1)=F(x+n)- \frac12.(\frac 1{x+n}-\frac 1{x+n+2})$, so $F(x+n+1) =F(x)-\frac12.(\frac 1x+\frac 1{x+1}-\frac 1{x+n+1}-\frac 1{x+n+2})$ Make $n\rightarrow \infty$ and you obtain $F(x)-\frac12.(\frac 1x+\frac 1{x+1})=C (constant)\Longrightarrow F(x)=\frac{2Cx^2+2(C+1)x+1)}{2x(x+1)}$ and so on ...

nice very cool solution

Two questions Diogene wrote: , so the equality holds for all $x\in R$, because of $P,Q$ are polynomials. Really? I can't quite see it. Diogene wrote: and you obtain $F(x)-\frac12.(\frac{1}{x}+\frac 1{x+1})=C (constant)$ Why constant?