Calculus: Early Transcendentals, Chapter 5, 5.4, Section 5.4, Problem 39

We will make substitution x=t^6. Therefore, the differential is dx=6t^5dt and the new bounds of integration are t_1=root(6)(1)=1 and t_2=root(6)(64)=2.
int_1^64(1+root(3)(x))/sqrt x dx=int_1^2((1+t^2)6t^5)/t^3dt=6int_1^2(1+t^2)t^2dt=6int_1^2(t^2+t^4)dt=
6(t^3/3+t^5/5)|_1^2=6(8/3+32/5-1/3-1/5)=6\cdot128/15=256/5 <-- The solution