y = x^4/8 + 1/(4x^2) , [1, 3] Find the arc length of the graph of the function over the indicated interval.

Arc length (L) of the function y=f(x) on the interval [a,b] is given by the formula,
L=int_a^b sqrt(1+(dy/dx)^2)dx , if y=f(x) a <=  x <=  b,
Now y=x^4/8+1/(4x^2)
Now we need to differentiate the above function with respect to x,
dy/dx=1/8(4)x^(4-1)+1/4(-2)x^(-2-1)
dy/dx=1/2x^3-1/2x^(-3)
dy/dx=x^3/2-1/(2x^3)
dy/dx=(x^6-1)/(2x^3)  
Now arc length L=int_1^3 sqrt(1+((x^6-1)/(2x^3))^2)dx
=int_1^3sqrt(1+(x^12-2x^6+1)/(4x^6))dx
=int_1^3sqrt((4x^6+x^12-2x^6+1)/(4x^6))dx
=int_1^3sqrt((x^12+2x^6+1)/(4x^6))dx
=int_1^3sqrt(((x^6+1)/(2x^3))^2)dx
=int_1^3(x^6+1)/(2x^3)dx
=int_1^3(x^6/(2x^3)+1/(2x^3))dx
=int_1^3(x^3/2+1/2x^(-3))dx
=[1/2x^4/4+1/2(x^(-3+1)/(-3+1))]_1^3
=[x^4/8-1/(4x^2)]_1^3
=[3^4/8-1/(4(3)^2)]-[1^4/8-1/(4(1)^2)]
=[81/8-1/36]-[1/8-1/4]
=[(729-2)/72]-[(1-2)/8]
=[727/72]-[-1/8]
=727/72+1/8
=(727+9)/72
=736/72
=92/9
So, the Arc length=92/9