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anonymous

  • 4 years ago

Need help integrating (x+1)/(x^3 + x).

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  1. slaaibak
    • 4 years ago
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    \[{x+1} \over {x(x^2 + 1)}\] Use partial fractions.

  2. anonymous
    • 4 years ago
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    I tried using partial fractions, and got A(x^2 + 1) + (Bx + C)x, but I'm not sure how to get the values of the coefficients.

  3. TuringTest
    • 4 years ago
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    \[A(x^2+1)+(Bx+C)x=x+1\to (A+B)x^2+Cx+A\]just looking at the coefficients it looks like A=1, C=1, and A+B=0 wich leads to B=-1

  4. slaaibak
    • 4 years ago
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    \[x + 1 = A(x^2 + 1) + (Bx + C) x\] \[x + 1 = (A+B)x^2 + Cx + A\] Therefore A + B = 0 C=1 A=1 damn turing beat me :/

  5. TuringTest
    • 4 years ago
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    sorry, you got your medal though ;)

  6. slaaibak
    • 4 years ago
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    We will continue this battle in another question ;)

  7. TuringTest
    • 4 years ago
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    indeed! did you get the idea Xylienda?

  8. anonymous
    • 4 years ago
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    :) Thanks guys!

  9. anonymous
    • 4 years ago
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    Yeah. So its more like a trial and error thing, right?

  10. TuringTest
    • 4 years ago
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    Not really in this case when you spell it out and look at the coefficient on the right side you get\[(A+B)x^2+Cx+A=x+1\]the coefficient of x^2 on the right is zero, so\[A+B=0\]the coefficient on x is 1, so\[C=1\]and the constant term is just 1, so\[A=1\]now that we know A=1 we can go back and solve the first one\[A+B=0\to B=-1\]so no, it's not trial and error. It is making a system from by setting the coefficients on each side equal. See?

  11. anonymous
    • 4 years ago
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    Oh! I think i really get it now. So you kind of begin by, more or less, matching up the forms on both sides of the equation and compare the coefficients, then work from there.

  12. TuringTest
    • 4 years ago
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    exactly collect like terms on the left and set each coefficient equal to that on the right. Sometimes the systems will be more complex, but here it is pretty easy.

  13. anonymous
    • 4 years ago
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    Thank you! :)

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