∫(0 2)dx/(1-x)^2求广义积分∫(0→2)dx/(1-x)^2

结果是发散的
∫(0，2) dx/(x-1)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)
= 1/(1-x) [0，1] + 1/(1-x) [1，2]
= [1/(1-1) - 1/(1-0)] + [1/(1-2) - 1/(1-1)]
= 1/0 - 1/0 - 2
= +∞

利用广义积分定义
∫(0→2)[1/(x-1)²]dx
=∫(0→1)[1/(x-1)²]dx+∫(1→2)[1/(x-1)²]dx
=∫(0→1)[1/(x-1)²]d(x-1)+∫(1→2)[1/(x-1)²]d(x-1)
=1/(1-x)|(0→1) + 1/(1-x)|(1→2)
={lim(x→1⁻)[1/(1-x)]-1/(1-0)}+{1/(1-2)-lim(x→1⁺)[1/(1-x)]}
={[+∞]-1/(1-0)}+{1/(1-2)-[-∞]}
=+∞-1-1+(+∞)
= +∞-2
=+∞

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).