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Thread: Another Partial Fraction Integration

  1. September 10th 2011, 11:35 PM #1
    celtic1234
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    Another Partial Fraction Integration

    Hi
    Can you tell me if i am correct here?
    thanks yet again
    John

    Using partial fractions Integrate

    \dfrac{(3x+1)}{(x^2)(2x+4)}

    \dfrac{(3x+1)}{(x^2)(2x+4)}=\frac{A}{(x^2)}+\frac{ B}{(2x+4)}....equation (1)

    Step 1:

    Multiply both sides by (x^2)(2x+4) gives;

    (3x+1)=(2x+4).A +(x^2).B

    Step 2:

    Make x=-2

    (3(-2)+1)=(2(-2)+4).A +((-2)^2).B gives;

    \frac {-5}{4}=B

    Step 3:

    Make x=0

    (3(0)+1)=(2(0)+4).A +((0)^2).B gives;

    \frac {1}{4}=A

    Step 4:

    Substitute into equation (1) for A & B

    \dfrac{(3x+1)}{(x^2)(2x+4)}=\frac{1}{4(x^2)}-\frac{5}{4(2x+4)}

    Step 5:

    now integrate the separate terms;

    \int \dfrac{(3x+1).dx}{(x^2)(2x+4)}=\int \frac{1.dx}{4(x^2)}-\int \frac{5.dx}{4(2x+4)}

    Step 6:

    =\frac{1}{4}\int \frac{dx}{(x^2)}-\frac{5}{4}\int \frac{dx}{(2x+4)}

    Step 7:

    =\frac{1}{4} ln(x^2) -\frac{5}{4}\frac{ln(2x+4)}{2} +C

    Step 8:

    =\frac{1}{4} ln(x^2) -\frac{5}{8}ln(2x+4) +C

    Step 9:

    =\frac{2}{8} ln(x^2) -\frac{5}{8}ln(2x+4) +C

    Step 10:

    =\frac{-3}{8} [ln(x^2) -ln(2x+4)] +C

    Step 11:

    =\frac{-3}{8} ln[(x^2) -(2x+4)] +C
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  2. September 10th 2011, 11:53 PM #2
    Siron
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    Re: Another Partial Fraction Integration

    The degree of the denominator has always to be greater (one degree) then the degree of the numerator, in this case you can write:
    \frac{3x+1}{x^2(2x+4)}=\frac{Ax+B}{x^2}+\frac{C}{2 x+4}
    \frac{(Ax+B)(2x+4)+Cx^2}{x^2(2x+4)}=\frac{2Ax^2+4A x+2Bx+4B+Cx^2}{x^2(2x+4)}
    =\frac{(2A+C)x^2+(4A+2B)x+4B}{x^2(2x+4)}

    Therefore:
    4A+2B=3 and 4B=1 \Leftrightarrow B=\frac{1}{4}
    And so 4A+\frac{1}{2}=3 \Leftrightarrow A=\frac{5}{8}
    And so 2A+C=0 \Leftrightarrow C=\frac{-5}{4}
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  3. September 11th 2011, 12:22 AM #3
    celtic1234
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    Re: Another Partial Fraction Integration

    thanks Siron
    I am a bit lost here?

    My book does not cover this at all!!

    Can you elaborate on "The degree of the denominator has always to be greater (one degree) then the degree of the numerator"

    Where is the \frac{(3x+1)}{x^2(2x+4)}=\frac{Ax+B}{x^2}+\frac{C} {(2x+4)} coming from?

    apologies if the is easy but i am lost
    appreciate the help if you have the time to explain it?
    John
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  4. September 11th 2011, 01:05 AM #4
    Siron
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    Re: Another Partial Fraction Integration

    I understand, but it's a little bit long to explain I think so I recommand you to do some research in the book you use. Try to look for some examples etc ... and afterwards try this exercice again .
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  5. September 11th 2011, 01:07 AM #5
    celtic1234
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    Re: Another Partial Fraction Integration

    ok will do
    thanks
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