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Math Help - Partial Fractions

  1. #1
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    Question Partial Fractions

    Express \frac{x^2 + 1}{x^2 - 1} in partial fractions and hence evaluate \int\frac{x^2 + 1}{x^2 - 1}  dx
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  2. #2
    TD!
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    First do the division, or use this trick:

    <br />
\frac{{x^2  + 1}}{{x^2  - 1}} = \frac{{x^2  - 1 + 2}}{{x^2  - 1}} = 1 + \frac{2}{{\left( {x - 1} \right)\left( {x + 1} \right)}}<br />

    Now do partial fractions on the last one.
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  3. #3
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    What do you mean by do partial fractions on the last one?
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by dadon
    What do you mean by do partial fractions on the last one?
    He means write:

    <br />
\frac{2}{(x-1)(x+1)}=\frac{A}{x-1}+\frac{B}{x+1}<br />

    Then:

    <br />
\frac{2}{(x-1)(x+1)}=\frac{A(x+1)+B(x-1)}{(x-1)(x+1)}=\frac{(A+B)x+(A-B)}{(x-1)(x+1)}<br />
.

    Hence:

    A+B=0, and A-B=2, which means A=1 and B=-1, and hence:

    <br />
\frac{2}{(x-1)(x+1)}=\frac{1}{x-1}-\frac{1}{x+1}.

    As each of the terms in this are easily to integrate your integral may be written:

    <br />
\int\frac{x^2 + 1}{x^2 - 1} dx=\int  1 + \frac{2}{{\left( {x - 1} \right)\left( {x + 1} \right)}}dx= \int \{1+\frac{1}{x-1}-\frac{1}{x+1}\} dx<br />
,

    which can be integrates easily.

    RonL
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  5. #5
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    Post Re:

    Thank you!

    You guys are certainly good at maths.

    Regards

    dadon
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