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    I am currently unable to integrate arcoshx, could someone please help?

    I started with:
    u = arcoshx
    du/dx = 1/ Square root(x^(2) -1)

    But then get stuck as it would be the integral of: u du * (Square root(x^(2) -1))

    Does anyone know where i went wrong and how to answer it?

    Thanks
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    (Original post by Davi6336)
    I am currently unable to integrate arcoshx, could someone please help?

    I started with:
    u = arcoshx
    du/dx = 1/ Square root(x^(2) -1)

    But then get stuck as it would be the integral of: u du * (Square root(x^(2) -1))

    Does anyone know where i went wrong and how to answer it?

    Thanks
    Integration by Parts I think
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    (Original post by M14B)
    Integration by Parts I think
    Alright sweet that worked, cheers
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    By parts, you should get x*arccosh(x) + (x+1)^1/2 * (x-1)^1/2 + C

    A trivial lapsus of my memory, I shouldn't rush things.
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    (Original post by Davi6336)
    Alright sweet that worked, cheers
    Not to be a buzzkill but I hope you understand why it worked, and the intuition behind finding that method. Many people don't.
    Basically, if you can integrate x times a function's derivative, then you can always integrate the function via IBP with 1 and the function itself.
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    (Original post by Bath~Student)
    By parts, you should get x*arccosh(x) + (1 - x^2)^1/2 + C
    False.
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    (Original post by Zacken)
    False.
    Hilarious. You have quite the sense of humour.
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    (Original post by Bath~Student)
    Hilarious. You have quite the sense of humour.
    No, it actually is false.
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    (Original post by Zacken)
    No, it actually is false.
    Fixed. Standard integrals and derivatives I memorise.. can happen.


    I shall redeem myself before I turn to my homework: it is an endeavour far more important.
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    (Original post by Bath~Student)
    Fixed.
    \sqrt{x+1}\sqrt{x-1} = \sqrt{x^2-1}, by the way.
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    (Original post by Zacken)
    \sqrt{x+1}\sqrt{x-1} = \sqrt{x^2-1}, by the way.
    Yes, and since RHS is prettier, why not keep it?

    I am right in the fixed integral, am I not?

    You are making me question reality. Perhaps my homework is what I should turn to..
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    (Original post by Bath~Student)
    I am right in the fixed integral, am I not?
    Yes.
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    (Original post by Zacken)
    Yes.
    I can breathe again.
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    (Original post by IrrationalRoot)
    Not to be a buzzkill but I hope you understand why it worked, and the intuition behind finding that method. Many people don't.
    Basically, if you can integrate x times a function's derivative, then you can always integrate the function via IBP with 1 and the function itself.
    Oh ok i see that point, because then it means the second part (the integral) of the integration by parts formula can be integrated. Cheers
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    (Original post by Davi6336)
    Oh ok i see that point, because then it means the second part (the integral) of the integration by parts formula can be integrated. Cheers
    Yep!
 
 
 
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