求暇积分【0,2】∫dx/(1-x)²
原式=【0,1】∫dx/(1-x)²+【1,2】∫dx/(1-x)²
=【0,1】∫dx/(x-1)²+【1,2】∫dx/(x-1)²
=【0,1】∫d(x-1)/(x-1)²+【1,2】∫d(x-1)/(x-1)²
=-1/(x-1)∣【0,1】-1/(x-1)∣【1,2】
=x→1⁻lim[-1/(x-1)]+1-{-1+x→1⁺lim[1/(x-1)]}
=x→1⁻lim[-1/(x-1)]+2-x→1⁺lim[1/(x-1)]=∞(发散).