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Math Help - Need help with Integration and completing the square

  1. #1
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    Need help with Integration and completing the square

    Integrate:

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

    I'm not really sure how to "complete the square"
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by plooms View Post
    Integrate:

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

    I'm not really sure how to "complete the square"
    Let z=x^{\frac{1}{2}}

    So we see that we have \frac{1}{(x^{\frac{1}{2}})^2+x^{\frac{1}{2}}}=\fra  c{1}{z^2+z}

    Can you complete the square now?

    --Chris
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  3. #3
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    Woah, thanks for that first step. I definitely wouldn't have looked at it that way.

    My main problem is I never learned how to "complete the square." I don't even know what that means!
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by plooms View Post
    Woah, thanks for that first step. I definitely wouldn't have looked at it that way.

    My main problem is I never learned how to "complete the square." I don't even know what that means!
    Let's say you had a quadratic equation of the form ax^2+bx=0

    To use the complete the square technique, the coefficient of x^2 must be 1.

    So we can rewrite the equation x^2+\frac{b}{a}x=0

    Now this is where we apply this technique.

    You need to introduce this new term into the equation.

    Take the coefficient of the x term, divide by 2, then square that quantity:

    \left(\frac{b}{2a}\right)^2

    In essence, we now add "zero" to the equation:

    x^2+\frac{b}{a}x+\left(\frac{b}{2a}\right)^2-\left(\frac{b}{2a}\right)^2=0.

    x^2+\frac{b}{a}x+\left(\frac{b}{2a}\right)^2 becomes a perfect square:

    x^2+\frac{b}{a}x+\left(\frac{b}{2a}\right)^2=\left  (x+\frac{b}{2a}\right)^2

    So our equation now takes on the general form \left(x+\frac{b}{2a}\right)^2-\left(\frac{b}{2a}\right)^2=0.

    We have "completed the square".

    Does this make sense?

    --Chris
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  5. #5
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    Sort of...

    I got lost on the last step before you generalized. Where did you get the x on the RHS?

    For my problem so far I've got:

    1/(z^2+z+(1/2)^2-(1/2)^2)
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  6. #6
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by plooms View Post
    Sort of...

    I got lost on the last step before you generalized. Where did you get the x on the RHS?
    Oh. That bit was just illustrating how that was a perfect square. That really wasn't the equation.

    For my problem so far I've got:

    1/(z^2+z+(1/2)^2-(1/2)^2)
    Yup.

    So that means that we should get \frac{1}{\left(z+\frac{1}{2}\right)^2-\frac{1}{4}}=\frac{4}{4\left(x+\frac{1}{2}\right)^  2-1}=\frac{4}{(2x+1)^2-1}

    --Chris
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  7. #7
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    Alternatively

    For \int \frac{1}{x^{1/2}+x} dx

    Taking x^1/2 common factor simplifies things:

    \int \frac{1}{x^{\frac{1}{2}}(1+x^{\frac{1}{2}})}dx

    u = 1+x^{\frac{1}{2}}
    2du = \frac{1}{x^{\frac{1}{2}}}dx
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  8. #8
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    alright thanks, i think i've got it from there
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