8x=2y^4+y^(-2)
implies x'=y^3- 1/4 y^(-3)
s = int_1^2 sqrt( 1 + (x')^2) \ dy
= int_1^2 sqrt( 1 + (y^3- 1/4 y^(-3))^2) \ dy
= int_1^2 sqrt( 1 + y^6 + 1/16 y^(-6)- 1/2) \ dy
= int_1^2 sqrt( y^6 + 1/16 y^(-6) + 1/2) \ dy
= int_1^2 sqrt( (y^3 + 1/4 y^(-3))^2) \ dy
= int_1^2 y^3 + 1/4 y^(-3) \ dy
= [ y^4/4 - 1/(8 y^2) ]_1^2
= 123/32