由 √x+√y=1,得 √y=1-√x,y'/(2√y)=-1/(2√x),则 y'=-√y/√x=-(1-√x)/√x=1-1/√x.L =∫√(1+y'^2)dx =∫√[1+(1-1/√x)^2]dx=∫√(2-2/√x+1/x)dx,令 √x=t,则 x=t^2,dx=2tdt,得L=∫√(2-2/t+1/t^2)2tdt = 2∫...