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Integral calculus

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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!

  • Integral calculus -

    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.

  • Integral calculus -

    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)

  • Integral calculus -

    Thank you both for taking your time to try this :)

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