Calculus of a Single Variable, Chapter 7, 7.4, Section 7.4, Problem 9

Arc length(L) of the function y=f(x) on the interval [a,b] is given by the formula,
L=int_a^bsqrt(1+(dy/dx)^2)dx , if y=f(x) and a<= x <= b
Now we have to differentiate the function,
y=x^5/10+1/(6x^3)
dy/dx=1/10(5)x^(5-1)+1/6(-3)x^(-3-1)
dy/dx=x^4/2-1/2x^(-4)
dy/dx=x^4/2-1/(2x^4)
dy/dx=1/2(x^4-1/x^4)
dy/dx=1/2((x^8-1)/x^4)
L=int_2^5sqrt(1+((x^8-1)/(2x^4))^2)dx
L=int_2^5sqrt(1+(x^16-2x^8+1)/(4x^8))dx
L=int_2^5sqrt((4x^8+x^16-2x^8+1)/(4x^8))dx
L=int_2^5sqrt((x^16+2x^8+1)/(4x^8))dx
L=int_2^5sqrt(((x^8+1)/(2x^4))^2)dx
L=int_2^5(x^8+1)/(2x^4)dx
L=int_2^5(x^8/(2x^4)+1/(2x^4))dx
L=int_2^5(x^4/2+1/(2x^4))dx
L=[1/2(x^(4+1)/(4+1))+1/2(x^(-4+1)/(-4+1))]_2^5
L=[x^5/10-1/(6x^3)]_2^5
L=[5^5/10-1/(6(5)^3)]-[2^5/10-1/(6(2)^3)]
L=[3125/10-1/750]-[32/10-1/48]
L=[(234375-1)/750]-[(768-5)/240]
L=[234374/750]-[763/240]
L=(1874992-19075)/6000
L=1855917/6000
L=309.3195