jmc

We have that $$\begin{array}{rl}f(f(x)+x)& =f(x^2+(a+1)x+b)\\ & =(x^2+(a+1)x+b{)}^2+a(x^2+(a+1)x+b)+b\\ & =x^4+(2a+2)x^3+(a^2+3a+2b+1)x^2+(a^2+2ab+a+2b)x+(ab+b^2+b).\end{array}$$We can write this as $$\begin{array}{rl}& x^4+(2a+2)x^3+(a^2+3a+2b+1)x^2+(a^2+2ab+a+2b)x+(ab+b^2+b)\\ & =x^2(x^2+ax+b)+(a+2)x^3+(a^2+3a+b+1)x^2+(a^2+2ab+a+2b)x+(ab+b^2+b)\\ & =x^2(x^2+ax+b)+(a+2)x\cdot (x^2+ax+b)+(a+b+1)x^2+(a^2+ab+a)x+(ab+b^2+b)\\ & =x^2(x^2+ax+b)+(a+2)x\cdot (x^2+ax+b)+(a+b+1)(x^2+ax+b)\\ & =(x^2+ax+b)(x^2+(a+2)x+(a+b+1)).\end{array}$$(The factor of $f(x)=x^2+ax+b$ should not be surprising. Why?)

Thus, we want $a$ and $b$ to satisfy $a+2=1776$ and $a+b+1=2010.$ Solving, we find $a=1774$ and $b=235,$ so $f(x)=\boxed{x^2+1774x+235}.$