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Thread: Integration of Rational Functions By Partial Fractions

  1. #1
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    Integration of Rational Functions By Partial Fractions

    I have this problem that needs to be solved by integration by partial fractions and I can't seem to get the right answer. Here's the problem as written:

    "Find $\displaystyle \int\frac{6x+1}{20x^2+14x-12}dx$ with respect to $\displaystyle x$, use the constant $\displaystyle C$."

    So, I have:

    $\displaystyle
    \int\frac{6x+1}{(5x+6)(4x-2)}dx = \int[\frac{A}{5x+6} + \frac{B}{4x-2}]dx
    $

    and


    $\displaystyle
    6x+1 = A(4x-2) + B(5x+6)
    $

    If $\displaystyle x=\frac{1}{2}$, then $\displaystyle B = \frac{8}{17}$

    and

    If $\displaystyle x = \frac{-6}{5}$, then $\displaystyle A = \frac{31}{850}$

    So,

    $\displaystyle
    \int[\frac{\frac{-41}{34}}{5x+6} + \frac{\frac{8}{17}}{4x-2}]dx$

    $\displaystyle
    = \frac{-41}{34}ln(5x+6) + \frac{8}{17}ln(4x-2) + C
    $

    Unfortunately, this is not the answer. I greatly appreciate if anyone can point out where I've gone wrong. I have a feeling it was a generally small, but ultimately large, error. It is entirely possible that I have made an arthimetic mistake, by the way.

    Thanks for your consideration,

    Austin Martin
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  2. #2
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    Quote Originally Posted by auslmar View Post
    I have this problem that needs to be solved by integration by partial fractions and I can't seem to get the right answer. Here's the problem as written:

    "Find $\displaystyle \int\frac{6x+1}{20x^2+14x-12}dx$ with respect to $\displaystyle x$, use the constant $\displaystyle C$."

    So, I have:

    $\displaystyle
    \int\frac{6x+1}{(5x+6)(4x-2)}dx = \int[\frac{A}{5x+6} + \frac{B}{4x-2}]dx
    $

    and


    $\displaystyle
    6x+1 = A(4x-2) + B(5x+6)
    $

    If $\displaystyle x=\frac{1}{2}$, then $\displaystyle B = \frac{8}{17}$

    and

    If $\displaystyle x = \frac{-6}{5}$, then $\displaystyle A = \frac{31}{850}$

    So,

    $\displaystyle
    \int[\frac{\frac{-41}{34}}{5x+6} + \frac{\frac{8}{17}}{4x-2}]dx$

    $\displaystyle
    = \frac{-41}{34}ln(5x+6) + \frac{8}{17}ln(4x-2) + C
    $

    Unfortunately, this is not the answer. I greatly appreciate if anyone can point out where I've gone wrong. I have a feeling it was a generally small, but ultimately large, error. It is entirely possible that I have made an arthimetic mistake, by the way.

    Thanks for your consideration,

    Austin Martin
    B is OK. A is wrong .... A = 31/34.
    Last edited by mr fantastic; Jun 22nd 2008 at 01:06 PM.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    B is correct. A is wrong .... A = 31/34.
    So the answer becomes

    $\displaystyle {\color{red}\frac{1}{5}} \, \frac{31}{34} \, \ln{\color{red}|} 5x + 6{\color{red}|} + {\color{red}\frac{1}{4}} \, \frac{{\color{red}8}}{17} \, \ln {\color{red}|} 4x - 2 {\color{red}|} + C = {\color{red}\frac{1}{5}} \, \frac{31}{34} \, \ln{\color{red} |} 5x + 6 {\color{red}|} + {\color{red}\frac{1}{4}} \, \frac{{\color{red}16}}{34} \, \ln {\color{red}|} 4x - 2 {\color{red}|} + C = .....$


    where you keep re-arranging until you recognise the correct answer.

    Note the use of absolute value!
    Last edited by mr fantastic; Jun 22nd 2008 at 03:16 PM. Reason: Careless careless careless of me .......
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  4. #4
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    Quote Originally Posted by auslmar View Post
    I have this problem that needs to be solved by integration by partial fractions and I can't seem to get the right answer. Here's the problem as written:

    "Find $\displaystyle \int\frac{6x+1}{20x^2+14x-12}dx$ with respect to $\displaystyle x$, use the constant $\displaystyle C$."

    So, I have:

    $\displaystyle
    \int\frac{6x+1}{(5x+6)(4x-2)}dx = \int[\frac{A}{5x+6} + \frac{B}{4x-2}]dx
    $

    and


    $\displaystyle
    6x+1 = A(4x-2) + B(5x+6)
    $

    If $\displaystyle x=\frac{1}{2}$, then $\displaystyle B = \frac{8}{17}$

    and

    If $\displaystyle x = \frac{-6}{5}$, then $\displaystyle A = \frac{31}{850}$

    So,

    $\displaystyle
    \int[\frac{\frac{-41}{34}}{5x+6} + \frac{\frac{8}{17}}{4x-2}]dx$

    $\displaystyle
    = \frac{-41}{34}ln(5x+6) + \frac{8}{17}ln(4x-2) + C
    $

    Unfortunately, this is not the answer. I greatly appreciate if anyone can point out where I've gone wrong. I have a feeling it was a generally small, but ultimately large, error. It is entirely possible that I have made an arthimetic mistake, by the way.

    Thanks for your consideration,

    Austin Martin
    If you need to I will go through the entire PFD for you, but a first thing that will make it a little bit easier. Why not finish your factorization?

    $\displaystyle 4x-2=2(2x-1)$

    So you can forego it
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  5. #5
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    Thanks to both of you. Now that it's established that I suck at arithmetic, I'm having trouble seeing why the answer is not:

    $\displaystyle
    \frac{31}{34}ln|5x+6| + \frac{8}{17}ln|4x-2| + C
    $ since $\displaystyle A = \frac{31}{34}$ and $\displaystyle B = \frac{8}{17}$
    Also Mr. Fantastic, I'm not sure what you mean by:


    Quote Originally Posted by mr fantastic View Post
    So the answer becomes

    $\displaystyle \frac{31}{34} \, \ln{\color{red}|} 5x + 6{\color{red}|} + \frac{4}{17} \, \ln {\color{red}|} 4x - 2 {\color{red}|} + C = \frac{31}{34} \, \ln{\color{red} |} 5x + 6 {\color{red}|} + \frac{8}{34} \, \ln {\color{red}|} 4x - 2 {\color{red}|} + C = .....$

    where you keep re-arranging until you recognise the correct answer.

    Note the use of absolute value!
    Re-arranging? Forgive my dense-ness.
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  6. #6
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    Quote Originally Posted by auslmar View Post
    Thanks to both of you. Now that it's established that I suck at arithmetic, I'm having trouble seeing why the answer is not:

    $\displaystyle
    \frac{31}{34}ln|5x+6| + \frac{8}{17}ln|4x-2| + C
    $ since $\displaystyle A = \frac{31}{34}$ and $\displaystyle B = \frac{8}{17}$

    Mr F says: That's one form of the correct answer.

    Also Mr. Fantastic, I'm not sure what you mean by:




    Re-arranging? Forgive my dense-ness.
    Other forms of the correct answer can be found by re-arranging the above. For example:

    $\displaystyle \frac{1}{34} \left( 31 \ln|5x+6| + 16 \ln|4x-2| + C \right)$.

    If you state what form the final answer is given in, you can be shown how to get that form from the above.
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  7. #7
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    Quote Originally Posted by mr fantastic View Post
    Other forms of the correct answer can be found by re-arranging the above. For example:

    $\displaystyle \frac{1}{34} \left( 31 \ln|5x+6| + 16 \ln|4x-2| + C \right)$.

    If you state what form the final answer is given in, you can be shown how to get that form from the above.
    Oh, I see. I use a homework system called Maple TA that allows the user to preview if their answer is correct. Unfortunately, when I enter any forms of the previously determined correct answer, the system says they're wrong. I can't figure out what's wrong. I'm using absolute values and including the costant $\displaystyle C$, so I'm puzzled as to why it's not recognizing the answer.
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  8. #8
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    Quote Originally Posted by mr fantastic View Post
    So the answer becomes

    $\displaystyle {\color{red}\frac{1}{5}} \, \frac{31}{34} \, \ln{\color{red}|} 5x + 6{\color{red}|} + {\color{red}\frac{1}{4}} \, \frac{4}{17} \, \ln {\color{red}|} 4x - 2 {\color{red}|} + C = {\color{red}\frac{1}{5}} \, \frac{31}{34} \, \ln{\color{red} |} 5x + 6 {\color{red}|} + {\color{red}\frac{1}{4}} \, \frac{{\color{red}16}}{34} \, \ln {\color{red}|} 4x - 2 {\color{red}|} + C = .....$

    where you keep re-arranging until you recognise the correct answer.

    Note the use of absolute value!
    My mistake. See the edit above. I was very careless. So were you, of course.

    In fact, if you follow mathstud's suggestion of simplifying 4x - 2 = 2(x - 1), you will probably get the form of the answer that's being looked for a lot easier ......

    Note that $\displaystyle \frac{1}{4} \, \frac{16}{34} \, \ln | 4x - 2 | + C = \frac{4}{34} \, \ln 2| 2x - 1 | + C$

    $\displaystyle = \frac{4}{34} \, \ln| 2x - 1 | + \frac{4}{34} \, \ln (2) + C $ where the last is just a constant, D say

    $\displaystyle = \frac{4}{34} \, \ln| 2x - 1 | + D = \frac{20}{170} \, \ln| 2x - 1 | + D $


    That should give you blue sky .........
    Last edited by mr fantastic; Jun 22nd 2008 at 03:21 PM. Reason: Incredible. Had to fix another careless mistake ..... Hmmpph! And moo promised to check all my posts ...... lol!
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