As Steve mentioned yesterday for this same question, the 5^(3/x) is a stopper.
I put your expression in the Wolfram integrator with the necessary changes in brackets, and frankly I don't even want to understand their answer.
http://integrals.wolfram.com/index.jsp?expr=%285%5E%283%2Fx%29+-+%28x%5E8+%2B+9%29%5E%281%2F3%29+%29+%2F+%283x%5E2%29&random=false
Yesterday I posted this number that I don't understand. There was a mistake in the question,this is the right one:
∫ (5^(3/x) - (x^8 + 9)^1/3 ) / 3x^2
will u, for sub. be = x^8 + 9?
Then du is = 8x^7 dx
x^7 dx = 1/8 du
Thank you!
3 answers
Actually, I have to reconsider.
If u = 5^(3/x)
du = ln5 5^(3/x) (-3/x^2)
we have the form, and we end up with
∫ (5^(3/x) - (x^8 + 9)^1/3 ) / 3x^2 dx
= -1/(9ln5) ∫ u - (x^8 + 9)^1/3 ) du
The show stopper is that ∛(x^8+9)
If u = 5^(3/x)
du = ln5 5^(3/x) (-3/x^2)
we have the form, and we end up with
∫ (5^(3/x) - (x^8 + 9)^1/3 ) / 3x^2 dx
= -1/(9ln5) ∫ u - (x^8 + 9)^1/3 ) du
The show stopper is that ∛(x^8+9)
Thank you both for taking your time to try this :)
0 answers