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unimatix

  • one year ago

Calculate the integral Integral x/((x-a)(x-b)) dx for a) a /= b (a does not equal b) b) a = b

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  1. Jhannybean
    • one year ago
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    Partial fraction decomposition?

  2. unimatix
    • one year ago
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    I know. I tried. But when I go to try and find anything I can't get anything. For example part a.|dw:1441687142216:dw|

  3. unimatix
    • one year ago
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    |dw:1441687204033:dw|

  4. unimatix
    • one year ago
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    |dw:1441687237178:dw|

  5. unimatix
    • one year ago
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    but that seems pretty useless

  6. UnkleRhaukus
    • one year ago
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    \[\begin{align} \frac1{(x-a)(x-b)}&=\frac A{x-a}+\frac B{x-b}\\ 1&=A(x-b)+B(x-a) \end{align}\] For \(x=a\) \[1=A(a-b)\implies A=\frac1{a-b}\] For \(x=b\) \[1=B(b-a)\implies B=\frac1{b-a}\] Therefore \[\begin{align} \frac1{(x-a)(x-b)}&=\frac 1{(a-b)(x-a)}+\frac 1{(b-a)(x-b)}\\[2ex] \frac x{(x-a)(x-b)}&=\frac x{(a-b)(x-a)}-\frac x{(a-b)(x-b)}\\ &=\frac1{a-b}\left(\frac x{x-a}-\frac x{x-b}\right)\\ \end{align}\]

  7. Jhannybean
    • one year ago
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    @UnkleRhaukus why is it \(\dfrac{1}{(x-a)(x-b)}\) instead of \(\dfrac{x}{(x-a)(x-b)}\)?

  8. zzr0ck3r
    • one year ago
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    You mean when he starts the decomp?

  9. Jhannybean
    • one year ago
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    Yeah.

  10. Jhannybean
    • one year ago
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    Just to make it easier to evaluate?

  11. zzr0ck3r
    • one year ago
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    Well notice that right after he says Therefor, he states the thing with 1 in the numerator, but then multiplies that result by x to get the desired equality

  12. zzr0ck3r
    • one year ago
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    yes, to make it easier :)

  13. Jhannybean
    • one year ago
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    Would it have changed the outcome in solving for \(A\) and \(B\) if he were to include \(x\) in the decomp process ?

  14. zzr0ck3r
    • one year ago
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    there would have been an x next to some things.

  15. zzr0ck3r
    • one year ago
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    exactly every \(1\) would be an \(x\), except the last one in the post :)

  16. Jhannybean
    • one year ago
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    That didnt make sense :\

  17. zzr0ck3r
    • one year ago
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    look at U.R. post, everywhere you see a \(1\)(except the last one) , replace is with an \(x\).

  18. Jhannybean
    • one year ago
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    Oh, yeah that makes sense.

  19. UnkleRhaukus
    • one year ago
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    \[\begin{align} \frac x{(x-a)(x-b)}&=\frac{Ax}{x-a}+\frac{Bx}{x-b}\\ 1&=Ax(x-b)+Bx(x-a) \end{align}\] For \(x=a\)\[1=Aa(a-b)\implies A=\frac1{a(a-b)}\] For \(x=b\)\[1=Bb(b-a)\implies B=\frac1{b(b-a)}\] Therefore: \[\begin{align} \frac x{(x-a)(x-b)} &= \frac x{a(a-b)(x-a)}+\frac x{b(b-a)(x-b)}\\ &= \frac1{a-b}\left(\frac x{x-a}-\frac x{x-b}\right) \end{align}\] @Jhannybean This is the same result; decomposing with the numerator term. In my opinion it is much harder to follow.

  20. Jhannybean
    • one year ago
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    Ok, yeah it does seem more complicated.

  21. UnkleRhaukus
    • one year ago
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    @unimatix, can you integrate now?

  22. unimatix
    • one year ago
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    Yep thanks for the help!

  23. Jhannybean
    • one year ago
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    @unimatix have you looked more into that other problem you were working on? Figured it out?

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