拆项法,应用公式1/[n(n+1)]=1/n-1/(n+1):
原式=1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+...+1/(x+2008)-1/(x+2009)
=1/(x+1)-1/(x+2009)
=2008/[(x+1)(x+2009)]