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Thread: [SOLVED] Integration by parts with a polynomial fraction inside?

  1. #1
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    [SOLVED] Integration by parts with a polynomial fraction inside?

    Here is what I have so far:
    <br />
\int ln(x^2+10x+16)~dx = xln(x^2+10x+16) - \int \frac {2x^2+10}{x^2+10x+16}~dx<br />

    I am stuck after this though...
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  2. #2
    Super Member 11rdc11's Avatar
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    long division of the polynomial
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  3. #3
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    How would this divide out? We have not really covered long division inside an integral in class yet. I know it splits into two integrals though right?
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  4. #4
    Super Member 11rdc11's Avatar
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    2 + \bigg(\frac{- 20x - 22}{x^2 +10x +16}\bigg)

    I would like to show you how to do the long divison but not sure how to display it on latex.
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  5. #5
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    Quote Originally Posted by redman223 View Post
    Here is what I have so far:
    <br />
\int ln(x^2+10x+16)~dx = x \ln(x^2+10x+16) - \int \frac {2x^2+10}{x^2+10x+16}~dx<br />

    I am stuck after this though...
    <br />
=x \ln(x^2+10x+16) - \int \frac {2x^2+20+32-10x-22}{x^2+10x+16}~dx<br />

    <br />
=x \ln(x^2+10x+16) - \int \left(2-\frac {10x+22}{x^2+10x+16}\right)~dx<br />

    =x \ln(x^2+10x+16) -2x + \int \frac {5(2x+10)-28} {x^2+10x+16} ~dx

    =x \ln(x^2+10x+16) -2x +5 \ln(x^2+10x+16) -28 \int \frac {1} {(x+5)^2-3^2} ~dx

    =x \ln(x^2+10x+16) -2x +5 \ln(x^2+10x+16) - \frac{28}{2\times 3} \ln \left( \frac {x+5+3} {x+5-3} \right)

    =x \ln(x^2+10x+16) -2x +5 \ln(x^2+10x+16) - \frac{14}{3} \ln \left( \frac {x+8} {x+2} \right)
    Last edited by Shyam; Sep 7th 2008 at 09:03 PM.
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  6. #6
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    there is a mistake in the first line written by redman: the numerator of the last integral should be x(2x+10), not 2x^2+10

    then \frac{2x^2+10x}{x^2+10x+16}=\frac{2(x^2+10x+16)-10x-32}{x^2+10x+16} and so on...
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  7. #7
    Super Member 11rdc11's Avatar
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    You could also use partial fractions or trig substitution, whatever you feel more comfortable with.
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by redman223 View Post
    Here is what I have so far:
    <br />
\int ln(x^2+10x+16)~dx = xln(x^2+10x+16) - \int \frac {2x^2+10}{x^2+10x+16}~dx<br />

    I am stuck after this though...
    actually, i'd approach it this way

    \int \ln (x^2 + 10x + 16)~dx = \int \ln [(x + 2)(x + 8)]~dx = \int \ln (x + 2) ~dx + \int \ln (x + 8)~dx

    Now in light of the fact that we know \int \ln x ~dx = x \ln x - x + C (we can get this via integration by parts, but you really should know this by heart), each of those two integrals are really easy with a simple linear substitution


    Haha, I just realized the date on this thing. really late, but it might help someone anyway....
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