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Math Help - Best way to integrate?

  1. #1
    Member billym's Avatar
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    Best way to integrate?

    What would be the best approach to integrating:

    x / (1 + x^2) , ( x > 0)

    (is "x arctan x + C" a possible solution?)

    By substituting u=(1+x^2) , is "ln sqrt (1 + x^2) + C" correct?
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  2. #2
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    Krizalid's Avatar
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    There's no way to get an arctangent there, it's just a hidden natural logarithm. Besides, your substitution works and your answer is correct.
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  3. #3
    Member billym's Avatar
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    Thanks alot.

    My problem then is:

    Given:

    ln sqrt (1+x^2) = y + 1/2(y^2) + C

    how would I solve for y?
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  4. #4
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    As far as I can see, you're probably solving a separable ODE.

    If you want to isolate y, calculus ends right here, it's just solving an algebra problem. You require to apply the quadratic formula. (Or may be completing the square could be helpful.)
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  5. #5
    Member billym's Avatar
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    Sorry, I'm trying to isolate x, not y.

    I have :

    x^2 = (e^(y + 1/2(y^2) + C)^2) - 1

    (let e^C = D)

    x = sqrt ((D*e^y*e(1/2(y^2))^2 - 1))

    Is this correct?
    Last edited by billym; February 18th 2008 at 10:41 AM.
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  6. #6
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    Deleted - I misread the equation.
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  7. #7
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    Quote Originally Posted by billym View Post
    Thanks alot.

    My problem then is:

    Given:

    ln sqrt (1+x^2) = y + 1/2(y^2) + C

    how would I solve for y?
    Quote Originally Posted by billym View Post
    Sorry, I'm trying to isolate x, not y.

    I have :

    x^2 = (e^(y + 1/2(y^2) + C)^2) - 1

    Mr F says:  \sqrt{1 + x^2} = e^{y + \frac{y^2}{2} + C} \Rightarrow 1 + x^2 = (e^{y + \frac{y^2}{2} + C})^2 = e^{2 \left( y + \frac{y^2}{2} + C \right)} = e^{2y + y^2 + 2C}


    (let e^C = D)

    Mr F says: No. Let e^{2C} = D. Therefore 1 + x^2 = D e^{2y + y^2} \Rightarrow x^2 = D e^{2y + y^2} - 1.

    And when you take the square root it'll be x = \pm ........ .


    x = sqrt ((D*e^y*e(1/2(y^2))^2 - 1))

    Is this correct?
    Note: (a^{m})^n = a^{mn}, NOT a^{m^n} ....
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  8. #8
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    And in fact the answer can be simplified a bit:

    x^2 = D e^{2y + y^2} - 1 = D e^{(y + 1)^2 - 1} - 1 = D \, e^{-1} e^{(y+1)^2} - 1.

    Let D\, e^{-1} = A. Then:

    x^2 = A \, e^{(y+1)^2} - 1.

    Therefore x = \pm \sqrt{A \, e^{(y+1)^2} - 1}.

    Alternatively, you can easily make y the subject if required .....
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