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

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 let's differentiate the function,
y=3/2x^(2/3)
dy/dx=3/2(2/3)x^(2/3-1)
dy/dx=1/x^(1/3)
Now let's plug the derivative in the arc length formula,
L=int_1^8sqrt(1+(1/x^(1/3))^2)dx
L=int_1^8sqrt(1+1/x^(2/3))dx
L=int_1^8sqrt((x^(2/3)+1)/x^(2/3))dx
L=int_1^8(1/x^(1/3))sqrt(x^(2/3)+1)dx
Now let's evaluate first the indefinite integral by using integral substitution,
Let t=x^(2/3)+1
dt=2/3x^(2/3-1)dx
dt/dx=2/(3x^(1/3))
dx/x^(1/3)=3/2dt
intsqrt(x^(2/3)+1)(1/x^(1/3))dx=int3/2sqrt(t)dt
=3/2(t^(1/2+1)/(1/2+1))
=3/2(t^(3/2)/(3/2))
=t^(3/2)
=(x^(2/3)+1)^(3/2)
L=[(x^(2/3)+1)^(3/2)]_1^8
L=[(8^(2/3)+1)^(3/2)]-[(1^(2/3)+1)^(3/2)]
L=[5^(3/2)]-[2^(3/2)]
L=11.18033989-2.828427125
L=8.351912763
Arc length (L) of the function over the given interval is ~~8.352