∫(x？ + 1)/(x+ 1)=x- 1+2/(x+ 1),∴(x？ + 1)/(x+ 1)-ax-b =( 1-a)x+2/(x+ 1)-(b+ 1).

Therefore, when x→∞, lim(x→∞)[(x? +1)/(x+1)-ax-b] = 0 holds, so 1-a=0, b+ 1=0.

∴a= 1、b=- 1。

For reference.