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anonymous

  • one year ago

Evaluate the integral (Partial Fractions) ∫x^2+1/(x-3)(x-2)^2 dx Why is it that when you split the function into A+B+C that you get A(x-2)+B(x-2)(x-3)+C(x-3). how come there isnt an A(x-2)^3

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  1. anonymous
    • one year ago
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    There's no A(x-2)^3 because (x-2) is only raised to the 2nd power. You don't need the second (x-3) either. \[\frac{ x^2+1 }{ (x-3)(x-2)^2 }=\frac{ A }{ x-3 }+\frac{ B }{ x-2 }+\frac{ C }{ (x-2)^2 }\]

  2. anonymous
    • one year ago
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    ok. How come there's C(x-3) and not C(x-2)(x-3). is it because there's a (x-2)^2 in the denominator?

  3. anonymous
    • one year ago
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    you mean when you multiply? You have to multiply by the least common denominator, so that's \((x-3)(x-2)^2\). So (x -3) cancels on the A leaving you with \(A(x-2)^2\) (x - 2) cancels on B, leaving \(B(x-3)(x-2)\) (x - 2)² cancels on C leaving \(C(x-3)\)

  4. anonymous
    • one year ago
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    \[x^2+1=A(x-2)^2+B(x-3)(x-2)+C(x-3)\]

  5. anonymous
    • one year ago
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    ok its makes a lot of sense now. thanks.

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