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FP2 Integral of arcoshx

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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
2. (Original post by Davi6336)

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
3. (Original post by M14B)
Integration by Parts I think
Alright sweet that worked, cheers
4. 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.
5. (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 times a function's derivative, then you can always integrate the function via IBP with 1 and the function itself.
6. (Original post by Bath~Student)
By parts, you should get x*arccosh(x) + (1 - x^2)^1/2 + C
False.
7. (Original post by Zacken)
False.
Hilarious. You have quite the sense of humour.
8. (Original post by Bath~Student)
Hilarious. You have quite the sense of humour.
No, it actually is false.
9. (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.
10. (Original post by Bath~Student)
Fixed.
, by the way.
11. (Original post by Zacken)
, 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..
12. (Original post by Bath~Student)
I am right in the fixed integral, am I not?
Yes.
13. (Original post by Zacken)
Yes.
I can breathe again.
14. (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 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
15. (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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