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Math Help - [SOLVED] Stuck on improper fractions in integrals

  1. #1
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    [SOLVED] Stuck on improper fractions in integrals

    I am trying to integrate:

    <br />
\int \frac {y^2}{1+25y^2}~dy<br />

    I am supposed to do long division right? I have come up with -1/25, but I don't know what to do with that number...
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  2. #2
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    One thing you could do is hammer it into a form that is recognizable.

    \int\frac{y^{2}}{1+25y^{2}}dy

    Let u=5y, \;\ \frac{du}{5}=dy

    When we make the subs we get:

    \frac{1}{125}\int\frac{u^{2}}{1+u^{2}}du

    \frac{1}{125}\left[\int du-\int\frac{1}{u^{2}+1}du\right]

    See the right integral?. Do you notice something that has to do with arctan?.

    Just integrate and resub the u=5y.
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  3. #3
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    \frac{y^2}{1+25y^2} = \frac{1}{25}\left(1 - \frac{1}{1+25y^2}\right)
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  4. #4
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    Equivalently: \int \frac{y^2}{1+25y^2} \ dy \ = \ \frac{1}{25} \int \frac{25y^2 {\color{red}+ 1 - 1}}{1 + 25y^2} \ dy \ = \ \frac{1}{25}\int \left(\frac{25y^2 + 1}{1+25y^2} - \frac{1}{1+(5y)^2}\right) \ dy
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  5. #5
    Super Member 11rdc11's Avatar
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    \int \frac{y^2}{1+25y^2}dy

    tan(u)=5y

    y=\frac{tan(u)}{5}

    dy=\frac{sec^2(u)}{5}du

    so \int \frac{y^2}{1+25y^2}dy = \int \frac {tan^2(u)}{125}du = \frac{1}{125} \bigg(\int sec^2(u)du - \int 1du\bigg)

    which is tan(u)-u



    I'll finish it up when I come back from tacobell, my girlfriend don't want to wait


    Ok came to finish \frac{1}{125}\bigg(5y - arctan(5y)\bigg)
    Last edited by 11rdc11; September 5th 2008 at 09:05 AM. Reason: Finishing up
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  6. #6
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    Ok thanks, this was just a small part of a larger problem, I just forgot what to do when it came to long division.

    The way skeeter did it is what we learned in class the other day, but I am confused as far as what to do next.

    <br />
\int \frac {1}{25} (1-\frac {1}{1+25y^2})<br />

    What do I do from here? Do I multiply the 1/25 by everything or do I pull the 1/25 out in front of the integral sign and create 2 more integrals like this:

    <br />
\frac {1}{25}\int 1 - \frac {1}{25}\int \frac {1}{1+25y^2}<br />
    Last edited by redman223; September 4th 2008 at 07:09 PM.
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  7. #7
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    5y-arctan(5y)

    Is this a solution to the original integral?
    <br />
\int \frac {y^2}{1+25y^2}~dy<br />

    I don't see exactly how that is the solution. I am stuck on the step before when you split the integral.
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  8. #8
    Moo
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    Quote Originally Posted by redman223 View Post
    Ok thanks, this was just a small part of a larger problem, I just forgot what to do when it came to long division.

    The way skeeter did it is what we learned in class the other day, but I am confused as far as what to do next.

    <br />
\int \frac {1}{25} (1-\frac {1}{1+25y^2})<br />

    What do I do from here? Do I multiply the 1/25 by everything or do I pull the 1/25 out in front of the integral sign and create 2 more integrals like this:

    <br />
\frac {1}{25}\int 1 - \frac {1}{25}\int \frac {1}{1+25y^2}<br />
    Please don't forget dy.

    \int \frac {1}{25} (1-\frac {1}{1+25y^2}) ~dy<br />

    Then you can pull out any constant you want, because the antiderivative of a*y is a*antiderivative of y.

    =\frac 1{25} \left(\int 1 ~dy - \int \frac{1}{1+25y^2} ~dy\right)=\frac 1{25} \int 1 ~dy-\frac 1{25} \int \frac{1}{1+25y^2} ~dy

    In the first one, calculate and then, divide by 25, properly.
    In the second one, substitute carefully t=5y, because you did it wrong..
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  9. #9
    Super Member 11rdc11's Avatar
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    oops i forgot multiply my final answer by 1/125
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