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  1. #1
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    Cool Integration

    Integrate (3+x)/(x^2 -9)^0.5

    For now, i would just like to know if i am supposed to use the by parts forumla?
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  2. #2
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    Quote Originally Posted by Erghhh View Post
    Integrate (3+x)/(x^2 -9)^0.5

    For now, i would just like to know if i am supposed to use the by parts forumla?
    \int \frac{x}{\sqrt{x^2-9}} \, dx + \int \frac{3}{\sqrt{x^2-9}} \, dx

    straight up substitution for the first integral , u = x^2 - 9

    trig sub for the second, x = 3\sec{t}
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  3. #3
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    Hello, Erghhh!

    I'd use a trig substitution . . .


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

    Substitute: . \int \frac{3\sec\theta + 3}{3\tan\theta}(3\sec\theta\tan\theta\,d\theta) \;=\;3\int(\sec\theta + 1)(\sec\theta\,d\theta)

    . . =\;3\int(\sec^2\!\theta + \sec\theta)\,d\theta \;=\;3\bigg[\tan\theta + \ln|\sec\theta + \tan\theta| \bigg] + C


    Back-substitute: . \sec\theta \:=\:\frac{x}{3} \:=\:\frac{hyp}{adj}
    . . Then: . opp = \sqrt{x^2-9} \quad\Rightarrow\quad \tan\theta \:=\:\frac{\sqrt{x^2-9}}{3}


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

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

    . . . . . . . = \;\sqrt{x^2-9} + 3\ln\left|x + \sqrt{x^2-9}\right| + C .**


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    ** .What happened to the denominator, you ask?
    . . . . Watch this!


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

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

    . . = \;\;\sqrt{x^2-9} \:+\: 3\ln\left|x + \sqrt{x^2-9}\right| \:-\: \underbrace{3\ln 3 \:+\: C}_{\text{a constant!}}

    . . = \;\;\sqrt{x^2-9} \:+\:3\ln\left|x + \sqrt{x^2-9}\right| + C


    Textbooks love to pull this on us . . . You've been warned!

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