一个高数问题中反常积分的敛散性:如果收敛，求其值。

I =∫& lt；1，+∞& gt；dx/[x√(2x^2-2x+1)=∫& lt；1，+∞& gt；dx/[x√[2(x-1/2)^2+1/2]

=(1/√2)∫& lt；1，+∞& gt；dx/[x√[(x-1/2)^2+1/4]

设x-1/2 = (1/2) tants，则√[(x-1/2)2+1/4]=(1/2)Sect，

x = (1/2)(1+tant)，dx = (1/2)(sect)^2 dt

I =(1/√2)∫& lt；π/4，π/2 & gt；(1/2)(sect)^2 dt/[(1/2)(1+tant)(1/2)Sect]

=√2∫& lt；π/4，π/2 & gt；secdt/(1+tant)=√2∫& lt；π/4，π/2 & gt；dt/(成本+烧结)

=∫& lt；π/4，π/2 & gt；dt/cos(t-π/4)=∫& lt；π/4，π/2 & gt；秒(t-π/4)d(t-π/4)

= { ln[sec(t-π/4)+tan(t-π/4)]} & lt；π/4，π/2 & gt；

= ln[sec(π/4)+tan(π/4)]-ln[sec 0+tan 0]

= ln(1+√2)-ln 1 = ln(1+√2)