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Dido525

  • one year ago

Someone want to explain this to me?

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  1. Dido525
    • one year ago
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  2. Dido525
    • one year ago
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    How the heck did wolfram jump from the first step to the second step? I feel like I am missing something.

  3. Dido525
    • one year ago
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    It's the second step I am not getting.

  4. Dido525
    • one year ago
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    \[\int\limits\limits_{}^{}\frac{ x }{ (x^2-2x+5)^2 }dx\] I put that intergal into wolfram. Then I put the show steps.

  5. Dido525
    • one year ago
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    THe picture I posted is near the final answer.

  6. Dido525
    • one year ago
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    I see no other way to solve it.

  7. Dido525
    • one year ago
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    Eww... More partial fractions?

  8. Dido525
    • one year ago
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    But we only have 1 term divedby a polynomial . You have to use partial fractions.

  9. Dido525
    • one year ago
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    Eww...

  10. Dido525
    • one year ago
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    Thanks anyways.

  11. Dido525
    • one year ago
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    Partial fractions do not help at all. You can't reduce this.

  12. Dido525
    • one year ago
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    Allright. It's almost 4:30 AM here. If someone could leave an explanation I would GREATLY appreciate it once I wake up in the morning.

  13. sirm3d
    • one year ago
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    \[\Large \sqrt{x^2-2x+5}=\sqrt{(x-1)^2 + 4}\] use trigonometric substitution \[\Large x-1=2\tan \theta\]

  14. Dido525
    • one year ago
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    @sirm3d : Okay why did you square root?

  15. sirm3d
    • one year ago
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    oh, sorry, it was squared. the partial fraction is \[\frac{x}{(x^2-2x+5)^2}=\frac{Ax+B}{x^2-2x+5}+\frac{Cx+D}{(x^2-2x+5)^2}\]

  16. Dido525
    • one year ago
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    So it's actually possible to split this up? O_o .

  17. TuringTest
    • one year ago
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    yep

  18. Dido525
    • one year ago
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    I got A,B,D = 0. C=1.

  19. Dido525
    • one year ago
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    Is that right>

  20. Dido525
    • one year ago
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    ?**

  21. Dido525
    • one year ago
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    @sirm3d

  22. sirm3d
    • one year ago
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    i see that, too. \[x=(1/2)(2x)=(1/2)(2x-2+2)=(1/2)(2x-2) + (1/2)(2)\]

  23. sirm3d
    • one year ago
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    \[\frac{x}{(x^2-2x+5)^2}=\frac{(1/2)(2x-2)}{(x^2-2x+5)^2}+\frac{(1/2)(2)}{(x^2-2x+5)^2}\] for the first term, use the substitution \[y=x^2-2x+5,\quad dy=(2x-2)dx\] for the second term, use the trigonometric substitution \[x-1=2\tan\theta\]

  24. Dido525
    • one year ago
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    So now I have: \[\int\limits_{}^{}\frac{ 1 }{ (x^2-2x+5)^2 }dx\]

  25. Dido525
    • one year ago
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    But we have a and b are 0 so the first term is entirely 0.

  26. sirm3d
    • one year ago
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    i realized that too, partial fraction doesn't work. check my previous post.

  27. Dido525
    • one year ago
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    \[x=Ax^3-2Ax^2+Bx^2-2Bx+Cx+5B+D\]

  28. Dido525
    • one year ago
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    Yeah okay hmm...

  29. Dido525
    • one year ago
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    x-1=2tan(theta) dx=2sec^2(theta) d(theta)

  30. Dido525
    • one year ago
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    @sirm3d

  31. Dido525
    • one year ago
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    \[\int\limits\limits_{}^{}\frac{ 1 }{ u^2 } du+ \int\limits\limits_{}^{}\frac{ 4\sec^2 \theta }{ 2(4\sec^2\theta) }d \theta\]

  32. Dido525
    • one year ago
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    I get that which doesn't seem right...

  33. Dido525
    • one year ago
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    I used u instead of y.

  34. sirm3d
    • one year ago
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    \[x-1=2\tan\theta\\x^2-2x+5=4\sec^2\theta\] \[\large (1/2)\int \frac{du}{u^2}+\int\frac{2\sec^2 \theta d\theta}{(2\sec \theta)^4}\]

  35. Dido525
    • one year ago
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    \[x-1=2\tan\theta\] \[dx=2\sec ^2\theta d \theta\]

  36. sirm3d
    • one year ago
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    yes. our numerators agree.

  37. sirm3d
    • one year ago
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    @amistre64 , can you take over? i have to catch some sleep.

  38. Dido525
    • one year ago
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    Thanks a lot :) .

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