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    Could anyone walk me through how to integrate the following please :

    2(x^2 + 3x - 1)/(x+1)(2x-1)

    thanks in advance
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    (Original post by guitarmike456)
    Could anyone walk me through how to integrate the following please :

    2(x^2 + 3x - 1)/(x+1)(2x-1)

    thanks in advance
    One approach would be to rewrite the fraction as

    \displaystyle \frac{(2x-1)(x+1) + ...}{(2x-1)(x+1)}

    and then use that to reach

    1 + \displaystyle \frac{A}{2x-1} + \frac{B}{x+1}

    where A and B are constants found in the usual way. This is then a simple integration.
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    (Original post by BJack)
    One approach would be to rewrite the fraction as

    \displaystyle \frac{(2x-1)(x+1) + ...}{(2x-1)(x+1)}

    and then use that to reach

    1 + \displaystyle \frac{A}{2x-1} + \frac{B}{x+1}

    where A and B are constants found in the usual way. This is then a simple integration.
    I tried that, and got A/x+1 + B/2x-1 = 5x + 1
    but that gives nasty numbers forB and A so I'm fairly sure its wrong?
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    2(x2+3x?1) / (x+1)(2x?1) ?1+A/x+1 +B /2x?1

    ? 2x2+6x?2?(x+1)(2x?1)+A(2x?1)+B(x +1)

    can someone confirm that the above is surely incorrect?
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    (Original post by guitarmike456)
    I tried that, and got A/x+1 + B/2x-1 = 5x + 1
    but that gives nasty numbers forB and A so I'm fairly sure its wrong?
    It should be 5x-1.
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    (Original post by BJack)
    It should be 5x-1.
    Thanks
 
 
 
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