令√(x-1)=t,则x=t^2+1,dx=2tdt,于是不定积分∫dx/[x√(x-1)] =∫2tdt/[(t^2+1)*t]=∫2dt/(t^2+1)=2tan^(-1) t+C=2tan^(-1) √(x-1)+C于是广义积分∫﹙2,﹢∞﹚dx/[x√﹙x-1﹚]=lim 2tan^(-1) √(x-1)-2*π/4=π-π/...