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Math Help - Partial Fractions

  1. #1
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    Partial Fractions

    If anyone could answer a question, I would appreciate it.

    I'm working on solving \int\frac{5-x}{2x^2+x-1} dx using partial fractions.

    Here's a summary of my work: \frac{5-x}{2x^2+x-1} = \frac{A}{2x-1}+\frac{B}{x+1} which leads to 5-x=A(x+1)+B(2x-1).

    Letting x = -1 yields -3B = 6, B = -2. Letting x = 1/2 yields 9/2 = 3/2A, or A = 3. This gives an integral with partial fractions: \int(\frac{3}{2x-1}-\frac{2}{x+1}) dx( which gives 3ln|2x-1|-2ln|x+1| + C.

    The text answer is (3/2)ln|2x-1|-2ln|x+1| + C. What am I missing?
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  2. #2
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    Quote Originally Posted by kaiser0792 View Post
    If anyone could answer a question, I would appreciate it.

    I'm working on solving \int\frac{5-x}{2x^2+x-1} dx using partial fractions.

    Here's a summary of my work: \frac{5-x}{2x^2+x-1} = \frac{A}{2x-1}+\frac{B}{x+1} which leads to 5-x=A(x+1)+B(2x-1).

    Letting x = -1 yields -3B = 6, B = -2. Letting x = 1/2 yields 9/2 = 3/2A, or A = 3. This gives an integral with partial fractions: \int(\frac{3}{2x-1}-\frac{2}{x+1}) dx( which gives 3ln|2x-1|-2ln|x+1| + C.

    The text answer is (3/2)ln|2x-1|-2ln|x+1| + C. What am I missing?

    Your partial fractions are OK but be caution :
    The integral  \int \frac{dx}{2x-1} is  \frac{1}{2}\ln|2x-1| + C . You have missed the  \frac{1}{2} .
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  3. #3
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    Thanks simplependulum, a careless mistake on my part, but your insight saved me much frustration.
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