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Math Help - Integration Challenge

  1. #1
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    Smile Integration Challenge

    Find the integral of (x^2+3)/(x^2+9)^1/2
    Last edited by nath_quam; June 18th 2006 at 03:36 AM.
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  2. #2
    Super Member malaygoel's Avatar
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    Quote Originally Posted by nath_quam
    Find the integral of (x^2+3)/(x^2+9)^1/2
    \int \frac {x^2+3}{\sqrt{x^2+9}}dx
    Change (x^2+3) into (x^2 +9) - 6 and you will get the answer.
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  3. #3
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    Hello, nath_quam!

    Anyway we do it, it's not simple . . .

    Find: \,\int \frac{x^2+3}{\sqrt{x^2+9}}\,dx
    Let x = 3\tan\theta\quad\Rightarrow\quad dx = 3\sec^2\!\theta\,d\theta

    Substitute: \int\frac{9\tan^2\!\theta + 3}{3\sec\theta}\,3\sec^2\!\theta\,d\theta \;= \;\int(9\tan^2\!\theta + 3)\sec\theta\,d\theta

    . . = \;\int[9(\sec^2\!\theta - 1) + 3]\sec\theta\,d\theta \;=\;\int(9\sec^2\!\theta - 6)\sec\theta\,d\theta<br />

    . . =\;9\int\sec^3\!\theta\,d\theta - 6\int\sec\theta\,d\theta


    The first integral must be done by parts (or a formula can be applied);

    . . \frac{9}{2}\bigg(\sec\theta\tan\theta + \ln|\sec\theta + \tan\theta|\bigg) - 6\ln\left|\sec\theta + \tan\theta\right| + C

    . . = \;\frac{9}{2}\sec\theta\tan\theta - \frac{3}{2}\ln\left|\sec\theta + \tan\theta\right| + C

    . . = \;\frac{3}{2}\bigg(3\sec\theta\tan\theta - \ln\left|\sec\theta + \tan\theta\right|\bigg) + C<br />
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  4. #4
    Super Member malaygoel's Avatar
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    Hello Soroban!
    How are you.
    Quote Originally Posted by Soroban
    Hello, nath_quam!

    Anyway we do it, it's not simple . . .


    Let x = 3\tan\theta\quad\Rightarrow\quad dx = 3\sec^2\!\theta\,d\theta

    Substitute: \int\frac{9\tan^2\!\theta + 3}{3\sec\theta}\,3\sec^2\!\theta\,d\theta \;= \;\int(9\tan^2\!\theta + 3)\sec\theta\,d\theta

    . . = \;\int[9(\sec^2\!\theta - 1) + 3]\sec\theta\,d\theta \;=\;\int(9\sec^2\!\theta - 6)\sec\theta\,d\theta<br />

    . . =\;9\int\sec^3\!\theta\,d\theta - 6\int\sec\theta\,d\theta


    The first integral must be done by parts (or a formula can be applied);

    . . \frac{9}{2}\bigg(\sec\theta\tan\theta + \ln|\sec\theta + \tan\theta|\bigg) - 6\ln\left|\sec\theta + \tan\theta\right| + C

    . . = \;\frac{9}{2}\sec\theta\tan\theta - \frac{3}{2}\ln\left|\sec\theta + \tan\theta\right| + C

    . . = \;\frac{3}{2}\bigg(3\sec\theta\tan\theta - \ln\left|\sec\theta + \tan\theta\right|\bigg) + C<br />
    Please give a final touch to your post(change theta into x in the answer you got).
    Keep Smiling
    Malay
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  5. #5
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    Hello, Malay!

    Please give a final touch to your post (change theta into x in the answer you got).

    I'd be happy to . . . I'll simplify it to the bitter, bitter end.


    We have: . \frac{3}{2}\left(3\sec\theta\tan\theta - \ln \left|\sec\theta + \tan\theta\right|\right) + C

    From x = 3\tan\theta, we have: . \tan\theta = \frac{x}{3}\quad\Rightarrow\quad\sec\theta = \frac{\sqrt{x^2+9}}{3}


    So we have: . \frac{3}{2}\left[\,3\cdot\frac{\sqrt{x^2+9}}{3}\cdot\frac{x}{3} - \ln\left|\frac{\sqrt{x^2+9}}{3} + \frac{x}{3}\right|\,\right] + C

    . . = \;\frac{3}{2}\left[\,\frac{x\sqrt{x^2+9}}{3} - \ln\left|\frac{x + \sqrt{x^2+9}}{3}\right|\,\right] + C

    . . = \;\frac{3}{2}\left[\,\frac{x\sqrt{x^2+9}}{3} - \left(\ln\left|x + \sqrt{x^2+9}\right| - \ln 3\right)\,\right] + C

    . . = \;\frac{3}{2}\left[\,\frac{x\sqrt{x^2+9}}{3} - \ln\left|x + \sqrt{x^2 + 9}\right| + \ln 3\,\right] + C

    . . = \;\frac{1}{2}x\sqrt{x^2+9} \:- \:\frac{3}{2}\ln\left|x + \sqrt{x^2+9}\right| \,+ \,\underbrace{\frac{3}{2}\ln 3 \,+ \,C}
    . . . . . . . . . . . . . . . . . . . . . . . . . \text{This is a constant}


    Therefore: . \frac{1}{2}x\sqrt{x^2+9} \:- \:\frac{3}{2}\ln\left|x + \sqrt{x^2+9}\right| \,+ \,C

    See what I mean?
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