Calculus Question (Integrals)?

Question

1. the integral of [2x^(5/3) - 9x] / [x^(1/3)] dx
I tried to do this by bringing the [x^(1/3)] up to the numerator as [x^(-1/3)] but im not sure what my U and Du would be after doing this.

2. the integral of e^(x)[6-e^(-x)]dx
I'm not sure where I should start out on this one

Answer 1

1. the integral of [2x^(5/3) - 9x] / [x^(1/3)] dx
=∫ 2x^{4/3} - 9x^{2/3} dx
=2x^{7/3} / (7/3) -9 x^{5/3}/(5/3) + C
= (6/7) x^{7/3} - ( 27/5)x^{5/3} + C /

2. the integral of e^(x)[6-e^(-x)]dx
=∫ 6e^x -1 dx
=6e^x -x + C .

Answer 2

1. Just divide each term in the numerator by the denominator and you will get:

∫[2x^(5/3) - 9x] / [x^(1/3)] dx = ∫[2x^(4/3) - 9x^(2/3)] dx

The rest should be easy.


2. Mulitply it out first.

∫e^(x)[6-e^(-x)]dx = ∫[6e^(x) -1]dx

Again, the rest should be easy.