y'=1/x,
s=∫ [√3,2√2] √ [ 1+(y')^2]dx
=∫ [√3,2√2] √ [1+(1/x)^2]dx,
先求其不定積分,然后再代入上下限,
令 x=cott.dx=-(csct)^2dt,
csct=√(1+x^2),
sint=1/√(1+x^2),
cost=x/√(1+x^2),
sett=√(1+x^2)/x,
=∫ √ [ 1+(1/x)^2]dx
=-∫ sect*(csct)^2dt
=-∫ [sect+sect*(cott)^2]dt
=-∫ sectdt-∫ costdt/(sint)^2
=-ln| sect+tant|-∫ dsint/(sint)^2
=-ln| sect+tant|+1/sint
=-ln|√(1+x^2)/x+1/x|+√(1+x^2)
∴原式=-[ln |3/(2√2)+1/(2√2)|-ln | √(1+3)/√3+1/√3|+√ (1+8)-√(1+3)
=1+(1/2)(ln3-ln2).