（1/x的二次方+x）+（2/x的二次方+4x+3）+（3/x的二次方+9x+18）
=1/[x(x+1)] + 2/[(x+1)(x+3)] + 3/[(x+3)(x+6)]
=[(1/x) - 1/(x+1)] + [1/(x+1) - 1/(x+3)] + [1/(x+3) - 1/(x+6)]
=1/x - 1/(x+6)
=6/[x(x+6)]