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Math Help - Integration formula problem

  1. #1
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    Integration formula problem

    Hi all,

    I'm having difficulty manipulating a problem so it fits the equation required to use the formula.

    I've been asked by using equation

    \int(f(x))^{n}f'(x)dx=\frac{1}{n+1}(f(x))^{n+1}+c
    show that

    \int\frac{e^{x}+e^{-2x}}{(2e^{x}-e^{-2x})^{\frac{3}{2}}}dx=-(2e^{x}-e^{-2x})^{-\frac{1}{2}}+c

    please don't reveal too much to me, but how can I manipulate

    \int\frac{e^{x}+e^{-2x}}{(2e^{x}-e^{-2x})^{\frac{3}{2}}}dx

    so it is of the form

    \int(f(x))^{n}f'(x)dx

    many thanks.
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  2. #2
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    Re: Integration formula problem

    Quote Originally Posted by Hydralisk View Post
    Hi all,

    I'm having difficulty manipulating a problem so it fits the equation required to use the formula.

    I've been asked by using equation

    \int(f(x))^{n}f'(x)dx=\frac{1}{n+1}(f(x))^{n+1}+c
    show that

    \int\frac{e^{x}+e^{-2x}}{(2e^{x}-e^{-2x})^{\frac{3}{2}}}dx=-(2e^{x}-e^{-2x})^{-\frac{1}{2}}+c

    please don't reveal too much to me, but how can I manipulate

    \int\frac{e^{x}+e^{-2x}}{(2e^{x}-e^{-2x})^{\frac{3}{2}}}dx

    so it is of the form

    \int(f(x))^{n}f'(x)dx

    many thanks.
    Firstly, what is the derivative of

    2e^x-e^{-2x}\;\;\;?
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  3. #3
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    Re: Integration formula problem

    Consider \displaystyle \frac{1}{f^{\frac{3}{2}}}= f^{\frac{-3}{2}} and \displaystyle\frac{-3}{2}+1 = \frac{-1}{2}
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  4. #4
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    Re: Integration formula problem

    I've got

    \frac{d}{dx}2e^{x}-e^{2x}=2e^{x}-2e^{2x}

    but unfortunately I can't see what to do with it now?
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  5. #5
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    Re: Integration formula problem

    Quote Originally Posted by pickslides View Post
    Consider \displaystyle \frac{1}{f^{\frac{3}{2}}}= f^{\frac{-3}{2}} and \displaystyle\frac{-3}{2}+1 = \frac{-1}{2}
    I'm sorry but this has gone "over my head", I think I may need a bit more "help an idiot" style tips.

    I feel a bit stupid asking for help, but I really can't seem to even make a start on it.
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  6. #6
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    Re: Integration formula problem

    Your derivative is incorrect. \displaystyle \frac{d}{dx}(2e^x - e^{-2x}) = 2e^x + 2e^{-2x} = 2(e^x + e^{-2x}).

    Rewrite your integral as

    \displaystyle \frac{1}{2}\int{2(e^{x} + e^{-2x})(2e^x - e^{-2x})^{-\frac{3}{2}}\,dx}
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  7. #7
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    Re: Integration formula problem

    Quote Originally Posted by pickslides View Post
    Consider \displaystyle \frac{1}{f^{\frac{3}{2}}}= f^{\frac{-3}{2}} and \displaystyle\frac{-3}{2}+1 = \frac{-1}{2}
    Does

    \int\frac{e^{x}+e^{-2x}}{(2e^{x}-e^{-2x})^{\frac{3}{2}}}dx=\int(e^{x}+e^{-2x})(2e^{x}-e^{-2x})^{-3/2}dx

    Am I on the right track? If I can get rid of the fraction I may be able to figure it out from there.
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  8. #8
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    Re: Integration formula problem

    Quote Originally Posted by Prove It View Post
    Your derivative is incorrect. \displaystyle \frac{d}{dx}(2e^x - e^{-2x}) = 2e^x + 2e^{-2x} = 2(e^x + e^{-2x}).

    Rewrite your integral as

    \displaystyle \frac{1}{2}\int{2(e^{x} + e^{-2x})(2e^x - e^{-2x})^{-\frac{3}{2}}\,dx}
    I "think" I may be OK from here on.

    Many many thanks,

    I've got to leave for work now, but tonight I'll give it a good try to carry on through it.

    Thanks again to all of you.
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  9. #9
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    Re: Integration formula problem

    Quote Originally Posted by Hydralisk View Post
    I've got

    \frac{d}{dx}\left(2e^{x}-e^{2x}\right)=2e^{x}-2e^{2x}

    but unfortunately I can't see what to do with it now?
    Yes, that derivative is correct,
    but you should have had a negative exponent on the 2nd term initially.
    Then you will see that this derivative is

    2\left(e^x+e^{-2x}\right)

    Then use

    \frac{1}{\left(2e^x-e^{-2x}\right)^{\frac{3}{2}}}=\frac{\left(2e^x-e^{-2x}\right)^0}{\left(2e^x-e^{-2x}\right)^{\frac{3}{2}}}

    =\left(2e^x-e^{-2x}\right)^{-\frac{3}{2}}
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  10. #10
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    Re: Integration formula problem

    Thank you all so much,

    I eventually got through it using everything you said above.
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