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Math Help - log vs inverse tan?????

  1. #1
    Member sluggerbroth's Avatar
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    log vs inverse tan?????

    integrate 1/((x)^(1/2)+9))

    books answer is 2x^(1/2)-log((x)^(1/2)+9) + c

    my question is can you also integrate this as 1/((x)^(1/4)^(2))+3^(2) to get (1/3) inverse tan x/3 + c????

    are these different versions of the same answer???
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: log vs inverse tan?????

    Quote Originally Posted by sluggerbroth View Post
    integrate 1/((x)^(1/2)+9))
    my question is can you also integrate this as 1/((x)^(1/4)^(2))+3^(2) to get (1/3) inverse tan x/3 + c????
    We have:

    \displaystyle\int \frac{dx}{x^{1/2}+9}=\displaystyle\int \frac{dx}{(x^{1/4})^2+3^2}=\frac{1}{9}\displaystyle\int \frac{dx}{\left(\frac{x^{1/4}}{3}\right)^2+1}

    But we need in the numerator the derivative of \frac{x^{1/4}}{3}.
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  3. #3
    Member sluggerbroth's Avatar
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    Re: log vs inverse tan?????

    are you saying (1/3)inverse tan x^(1/4)/3 is the same as books answer
    of 2x^(1/2)-log((x)^(1/2)+9) + c
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    Re: log vs inverse tan?????

    No you cannot use inverse tan here... the power at the bottom is 1/2, NOT 2.

    You have

    \frac{1}{\sqrt{x}+9}

    Make the substitution x = u^2. This will lead you to a quotient of two linear equations; simplify them using polynomial division, and then the result should be easy to integrate.
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  5. #5
    Member sluggerbroth's Avatar
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    Re: log vs inverse tan?????

    If you re wrote this as 1/ (((x)^(1/4))^(2)+3^(2))

    i see how it works your way but why would the way have it above NOT WORK? It looks just like the integral of 1/(a^(2)+U^(2))=1/a (inverse tan u/a) +c

    what am i missing??/
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  6. #6
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    Re: log vs inverse tan?????

    No because you need the derivative of the inside to apply the chain rule.

    For example if

    f(x) = \tan^{-1}({3x^2}) then

    f'(x) = \frac{6x}{1+(3x^2)^2}

    But in our case the derivative of the inside isn't available so inverse tan won't work..
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