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Math Help - Completing the perfect square and tables of integration

  1. #1
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    Completing the perfect square and tables of integration

    I had to Complete the perfect square and use the tables of integration to integrate the following:
    ∫ 1 / (√x^2 3x + 6) dx

    What I have done:
    Completing the perfect square we have;
    x^2 3x + 6 = x^2 3x + 2.25 + 6 - 2.25 = (x 1.5)^2 +3.75
    if we set u = x 1.5, then du = dx, the integral in terms of u is
    ∫ 1 / √(u^2 2.25) du

    We then have
    ∫ 1 / √(u^2 2.25) du = ln|u +√u^2 2.25|,

    And the answer in terms of x is
    ln|x + 2 +√(u^2 2.25)| + C

    Have I done this correctly?
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  2. #2
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    Quote Originally Posted by mellowdano View Post
    I had to Complete the perfect square and use the tables of integration to integrate the following:
    ∫ 1 / (√x^2 3x + 6) dx

    What I have done:
    Completing the perfect square we have;
    x^2 3x + 6 = x^2 3x + 2.25 + 6 - 2.25 = (x 1.5)^2 +3.75
    if we set u = x 1.5, then du = dx, the integral in terms of u is
    ∫ 1 / √(u^2 2.25) du

    We then have
    ∫ 1 / √(u^2 2.25) du = ln|u +√u^2 2.25|,

    And the answer in terms of x is
    ln|x + 2 +√(u^2 2.25)| + C

    Have I done this correctly?
    Shouldn't the integral become after substitution

     <br />
\int \frac{1}{\sqrt{u^2 + 3.75}}\,du<br />
?
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  3. #3
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    Krizalid's Avatar
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    the way i'd complete the square is the following:

    \int{\frac{dx}{\sqrt{{{x}^{2}}-3x+6}}}=2\int{\frac{dx}{\sqrt{4{{x}^{2}}-12x+24}}}=2\int{\frac{dx}{\sqrt{{{(2x-3)}^{2}}+15}}}, now put 2x-3=\sqrt{15}t.
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  4. #4
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    Quote Originally Posted by Danny View Post
    Shouldn't the integral become after substitution

     <br />
\int \frac{1}{\sqrt{u^2 + 3.75}}\,du<br />
?
    yes your probably right,

    so by making those changes i would have

    Completing the perfect square we have;
    x^2 3x + 6 = x^2 3x + 2.25 + 6 - 2.25 = (x 1.5)^2 +3.75
    if we set u = x 1.5, then du = dx, the integral in terms of u is
    ∫ 1 / √(u^2 + 3.75) du

    We then have
    ∫ 1 / √(u^2 + 3.75) du = ln|u +√u^2 + 3.75|,

    And the answer in terms of x is
    ln|x + 2 +√(u^2 + 3.75)| + C
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  5. #5
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    I haven't learnt to complete square the way you have, I am trying to follow a similar example in my book. I have posted a correction that Danny pointed out.
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