`int_1^64 (1 + root(3)(x))/sqrt(x) dx` Evaluate the integral

Textbook Question

Chapter 5, 5.4 - Problem 39 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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tiburtius | High School Teacher | (Level 2) Educator

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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

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