求这道微积分的解法 

∫ (1->√3) √ ( 1 + 1/x^2) dx
=∫ (1->√3) (1/x)√ ( x^2 + 1) dx
let
x= tany
dx = (secy)^2 dy
x=1,y=π/4
x= √3,y= π/3
∫ (1->√3) √ ( 1 + 1/x^2) dx
= ∫ (π/4->π/3) (1/tany) (secy)^3 dy
= ∫ (π/4->π/3) (cscy) dtany
= (cscy.tany)|(π/4->π/3) +∫ (π/4->π/3) cscy dy
=(2 -√2) + (ln|cscy - coty|)| (π/4->π/3)
=(2 -√2) + ln(1/√3) - ln(√2 -1)
=(2 -√2) -(1/2)ln3 - ln(√2 -1)