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  1. #1
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    need to solve integration

    need to solve integration-316507_269396843094071_100000712735194_877144_1660221252_n-1-.jpg
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  2. #2
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    Re: need to solve integration

    Quote Originally Posted by Mhmh96 View Post
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    Where does this "need" come from? Is that the original question? If not please post the real/original question.

    CB
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  3. #3
    MHF Contributor Amer's Avatar
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    Re: need to solve integration

    t=\sqrt[x]{2\sqrt[x]{2\sqrt[x]{2...}}}

    t=\sqrt[x]{2t}\Rightarrow t^x = 2t \Rightarrow t = 2^{\frac{1}{1-x}}

    2^{\frac{1}{1-x}} = e^{ln(2^{\frac{1}{1-x}})}= e^{\frac{1}{1-x}ln(2)}
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  4. #4
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    Re: need to solve integration

    The answer that gave to me here is not relevent to my problem

    I need answer please !
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    Re: need to solve integration

    Thank you Amer i did not see your answer ^
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  6. #6
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    Re: need to solve integration

    Amer's advice in post 4 gives you \displaystyle \sqrt[x]{2\sqrt[x]{2\sqrt[x]{2\sqrt[x]{\dots}}}} = e^{\frac{\ln{2}}{1 - x}}, so

    \displaystyle \begin{align*} \int{\frac{\sqrt[x]{2\sqrt[x]{2\sqrt[x]{2\sqrt[x]{\dots}}}}}{(1 - x)^2}\,dx} &= \int{\frac{e^{\frac{\ln{2}}{1 - x}}}{(1 - x)^2}\,dx} \\ &= \frac{1}{\ln{2}}\int{e^{\frac{\ln{2}}{1 - x}}\left[\frac{\ln{2}}{(1 - x)^2}\right]\,dx} \end{align*}

    Which now can be solved using the substitution \displaystyle u = \frac{\ln{2}}{1 - x}.

    In future I suggest you read and try to understand and use the help that you are given instead of telling these people that their advice is "not relevant" and demanding an answer.
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    MHF Contributor chisigma's Avatar
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    Re: need to solve integration

    Quote Originally Posted by Mhmh96 View Post
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    The 'core' of the question is the computation of...

    f(x)= \lim_{n \rightarrow \infty} a_{n} (1)

    ... where a_{n} satisfies the difference equation...

    a_{n+1}= (2\ a_{n})^{\frac{1}{x}} (2)

    The limit (1), if a finite limit exists, is the solution of the equation...

    \Delta_{n}= a_{n+1}-a_{n}= f(a_{n})=0 (3)

    ... where...

    f(\xi)= (2\ \xi)^{\frac{1}{x}}-\xi (4)

    With a little of patience You find that the solution of (3) is...

    \xi= 2^{- \frac{1}{1-x}} (5)

    ... so that is...

    \int \frac{2^{- \frac{1}{1-x}}}{(1-x)^{2}}\ dx = \frac{1}{\ln 2} 2^{-\frac{1}{1-x}} + c (6)

    Kind regards

    \chi \sigma
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  8. #8
    MHF Contributor Amer's Avatar
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    Re: need to solve integration

    Quote Originally Posted by Amer View Post
    t=\sqrt[x]{2\sqrt[x]{2\sqrt[x]{2...}}}

    t=\sqrt[x]{2t}\Rightarrow t^x = 2t \Rightarrow t = 2^{\frac{1}{1-x}}

    2^{\frac{1}{1-x}} = e^{ln(2^{\frac{1}{1-x}})}= e^{\frac{1}{1-x}ln(2)}
    i made a mistake there is a minus sign 2^{\frac{-1}{1-x}} = e^{ln(2^{\frac{-1}{1-x}})}= e^{\frac{-1}{1-x}ln(2)}
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  9. #9
    A Plied Mathematician
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    Re: need to solve integration

    Quote Originally Posted by Mhmh96 View Post
    The answer that gave to me here is not relevent to my problem

    I need answer please !
    Quote Originally Posted by Prove It View Post
    Amer's advice in post 4 gives you \displaystyle \sqrt[x]{2\sqrt[x]{2\sqrt[x]{2\sqrt[x]{\dots}}}} = e^{\frac{\ln{2}}{1 - x}}, so

    \displaystyle \begin{align*} \int{\frac{\sqrt[x]{2\sqrt[x]{2\sqrt[x]{2\sqrt[x]{\dots}}}}}{(1 - x)^2}\,dx} &= \int{\frac{e^{\frac{\ln{2}}{1 - x}}}{(1 - x)^2}\,dx} \\ &= \frac{1}{\ln{2}}\int{e^{\frac{\ln{2}}{1 - x}}\left[\frac{\ln{2}}{(1 - x)^2}\right]\,dx} \end{align*}

    Which now can be solved using the substitution \displaystyle u = \frac{\ln{2}}{1 - x}.

    In future I suggest you read and try to understand and use the help that you are given instead of telling these people that their advice is "not relevant" and demanding an answer.
    The OP'er was likely referring to some deleted posts of chisigma's.
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  10. #10
    MHF Contributor chisigma's Avatar
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    Re: need to solve integration

    Quote Originally Posted by Ackbeet View Post
    The OP'er was likely referring to some deleted posts of chisigma's.
    I confirm that!... my first approach was fully wrong and I apologize for that!...

    Kind regards

    \chi \sigma
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