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Math Help - How to integrate 1/(y^2+3y) dy?

  1. #1
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    How to integrate 1/(y^2+3y) dy?

    1
    ---------- dy
    y^2 + 3y


    Please help!!
    Last edited by orange890; May 18th 2006 at 08:41 AM. Reason: missing something
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by orange890
    1
    ---------- dy
    y^2 + 3y


    Please help!!
    How do you find:

    <br />
\int \frac{1}{y^2+3y} \ dy<br />
?

    You resolve the integrand into partial fractions, which can then be integrated
    by inspection.

    <br />
\frac{1}{y^2+3y}=\frac{1}{y(y+3)}=\frac{A}{y}+ \frac{B}{y+3}<br />

    With a bit of manipulation you will find:

    <br />
\frac{1}{y^2+3y}=\frac{1}{3}\left\{ \frac{1}{y}- \frac{1}{y+3} \right\}<br />
.

    You should now be able to find the integral yourself from this point.

    RonL
    Last edited by CaptainBlack; May 18th 2006 at 11:12 AM. Reason: correct wrong signs
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  3. #3
    TD!
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    Quote Originally Posted by CaptainBlack
    With a bit of manipulation you will find:

    <br />
\frac{1}{y^2+3y}=\frac{1}{3}\left\{ \frac{-1}{y}+ \frac{1}{y+3} \right\}<br />
.
    I believe the signs should be switched, so:

    <br />
\frac{1}{y^2+3y}=\frac{1}{3}\left\{ \frac{1}{y}+ \frac{-1}{y+3} \right\}<br />
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by TD!
    I believe the signs should be switched, so:

    <br />
\frac{1}{y^2+3y}=\frac{1}{3}\left\{ \frac{1}{y}+ \frac{-1}{y+3} \right\}<br />
    So it does

    RonL
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  5. #5
    TD!
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    Don't be angry, be happy. Making mistakes is a sign of being (only?) human
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by TD!
    Don't be angry, be happy. Making mistakes is a sign of being (only?) human
    Hey, I once had a Mechanics prof who would take 15 points off for missing a "-" sign. (Yes, it WAS out of a 100 point exam.) The bad news was that his lectures contained just those sorts of errors. (We were all happy to get B's in that class!)

    -Dan
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