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    I've been working on
     \int \frac {xtan^{-1}xdx}{ \sqrt {1+x^2} }
    Integrating by parts, I've come across
     \sqrt {1+x^{2}} tan^{-1}x - \int \frac {dx}{ \sqrt {1+x^{2}}}
    Which according to wolframalpha is
     \sqrt {1+x^{2}}tan^{-1}x - sinh^{-1}x
    However I can't prove it, so:
    Could anyone of you prove to me why
     \int \frac {dx}{ \sqrt {1+x^{2}}} = sinh^{-1}x
    Or better, help me, myself, jump to that conclusion

    It may be a bit lame, as I have never been properly introduced to hyperbolic functions. I only now what sinhx, and coshx are.

    Looking hopefully to your replies ^^
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    cause it's in your formula book?
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    (Original post by Papkin)
    I've been working on
     \int \frac {xtan^{-1}xdx}{ \sqrt {1+x^2} }
    Integrating by parts, I've come across
     \sqrt {1+x^{2}} tan^{-1}x - \int \frac {dx}{1+ \sqrt {1+x^{2}}}
    Which according to wolframalpha is
     \sqrt {1+x^{2}}tan^{-1}x - sinh^{-1}x
    However I can't prove it, so:
    Could anyone of you prove to me why
     \int \frac {dx}{ \sqrt {1+x^{2}}} = sinh^{-1}x
    Or better, help me, myself, jump to that conclusion

    It may be a bit lame, as I have never been properly introduced to hyperbolic functions. I only now what sinhx, and coshx are.

    Looking hopefully to your replies ^^
    x=\tan u would have been better from the start... Then tidy it up and use IBP. No need for hyperbolics for this integral. If you want the explanation about arcsinh, read Mr M's post.
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    (Original post by Papkin)
    Could anyone of you prove to me why
     \int \frac {dx}{ \sqrt {1+x^{2}}} = sinh^{-1}x
    Or better, help me, myself, jump to that conclusion
    y = \sinh ^{-1} x

    x = \sinh y

    \frac{dx}{dy} = \cosh y

    \cosh^2 y = 1 + \sinh^2 y

    Can you form an expression for \frac{dy}{dx} in terms of x?

    Remember cosh is a positive function.
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    Substitution. Try substituting x with sinhy
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    (Original post by Mr M)
    \cosh^2 y = 1 + \sinh^2 y
    .
    Yep, that was lame, I did not know that^^
    Anyway, thanks for your help.

    (Original post by Farhan.Hanif93)
    x=\tan u would have been better from the start... Then tidy it up and use IBP. No need for hyperbolics for this integral. If you want the explanation about arcsinh, read Mr M's post.
    How is it relevant?
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    (Original post by Papkin)
    Yep, that was lame, I did not know that^^
    You might like to prove it by using the exponential definitions of sinh and cosh.

    If you are interested, Google Osborne's Rule.

    http://everything2.com/title/Osborne%2527s+rule
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    (Original post by Papkin)
    Yep, that was lame, I did not know that^^
    Anyway, thanks for your help.


    How is it relevant?
    I=\displaystyle\int \dfrac{x\tan ^{-1}x}{\sqrt{1+x^2}}dx
    x=\tan u \implies dx=\sec ^2u du
    Therefore:
    I=\displaystyle\int \dfrac{u\tan u\sec ^2u}{\sqrt{1+\tan ^2u}}du
    \implies \displaystyle\int u\tan u \sec u du.
    Now do IBP with p=u and \frac{dq}{du}=\tan u \sec u.
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    (Original post by Mr M)
    You might like to prove it by using the exponential definitions of sinh and cosh.

    If you are interested, Google Osborne's Rule.

    http://everything2.com/title/Osborne%2527s+rule
    Thanks, done that. Also, the Osborne's rule is pretty cool.
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    (Original post by Farhan.Hanif93)
    I=\displaystyle\int \dfrac{x\tan ^{-1}x}{\sqrt{1+x^2}}dx
    x=\tan u \implies dx=\sec ^2u du
    Therefore:
    I=\displaystyle\int \dfrac{u\tan u\sec ^2u}{\sqrt{1+\tan ^2u}}du
    \implies \displaystyle\int u\tan u \sec u du.
    Now do IBP with p=u and \frac{dq}{du}=\tan u \sec u.
    I didn't imply it was wrong. I implied it was irrelevant to the question I asked.
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    (Original post by Papkin)
    I didn't imply it was wrong. I implied it was irrelevant to the question I asked.
    Apologies, I offered you an alternative path since you implied you weren't comfortable with a fact about hyperbolics... That was all.
 
 
 
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