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chrisplusian Group Title

Quick integration question

  • 2 years ago
  • 2 years ago

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  1. chrisplusian Group Title
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    \[\int\limits x-1+\frac{ 1 }{ x+2 } +\frac{ 1 }{ x-1 } dx\]

    • 2 years ago
  2. chrisplusian Group Title
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    is this the same as:

    • 2 years ago
  3. chrisplusian Group Title
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    \[\int\limits x-1 dx +\int\limits \frac{ 1 }{ x+2 } dx +\int\limits \frac{ 1 }{ x-1 } dx\] ????

    • 2 years ago
  4. AccessDenied Group Title
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    Yes, that would be correct.

    • 2 years ago
  5. chrisplusian Group Title
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    I am doing integration by partial fraction decomposition, I got this far on my own but when I checked wolfram alpha it showed the "x-1" as the integral of x dx, and then the integral of -1 dx and somehow their answer and my own do not match

    • 2 years ago
  6. AccessDenied Group Title
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    You may also split (x - 1) into separate integrals, that should work out as well.

    • 2 years ago
  7. chrisplusian Group Title
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    are these answers the same:

    • 2 years ago
  8. chrisplusian Group Title
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    \[x^2 -2x +1 +2\ln \left| x+2 \right| +2\ln \left| x-1 \right|\]

    • 2 years ago
  9. chrisplusian Group Title
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    and

    • 2 years ago
  10. chrisplusian Group Title
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    \[\frac{ 1 }{ 2 }(x-2)x +\ln (x-1) +\ln (x+2)\]

    • 2 years ago
  11. chrisplusian Group Title
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    ??

    • 2 years ago
  12. chrisplusian Group Title
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    The first one is my answer the second is wolfram alpha

    • 2 years ago
  13. AccessDenied Group Title
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    Your answer appears to be about the same as wolfram's, except it was multiplied by 2, which doesn't work. The absolute values are fine to have here. Also, since you are doing an indefinite integration, both answers would have an additional arbitrary constant.

    • 2 years ago
  14. chrisplusian Group Title
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    when I had the integral of x-1 dx I substituted and said u=x-1 and du=dx. then integrating I ended up with (u^2)/2. plugging back in for u I got the polynomial over two. to clear the fraction from the entire thing I multiplied each term by 2 hence the 2ln.... is that wrong?

    • 2 years ago
  15. AccessDenied Group Title
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    yeah, that wouldn't work out because you essentially changed the value of the original integral. Multiplying by 2 also multiplies your original integral all by 2.

    • 2 years ago
  16. chrisplusian Group Title
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    I can see how my answer is similar to wolfram's but, when you distribute the x into the (x-2) of wolframs solution you get (x^2)-2x an that is all divided by two. My solution was (X^2)-2x +1.

    • 2 years ago
  17. chrisplusian Group Title
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    so you are saying that I can't take\[\frac{ x^2 -2x +1 }{ 2 } +\ln \left| x+2 \right| +\ln \left| x-1 \right| +c\]

    • 2 years ago
  18. chrisplusian Group Title
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    and multiply through by two to clear the fraction?

    • 2 years ago
  19. AccessDenied Group Title
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    Yes, since you basically have: \[ \int (x - 1 + \frac{1}{x+2} + \frac{1}{x-2}) \; dx = \frac{1}{2}(x^2 - 2x + 1) + \ln|x + 2| + \ln|x-1| + c \] You'd have to multiply both sides by 2 for the equality to hold.

    • 2 years ago
  20. AccessDenied Group Title
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    So, you'd have a slightly different integral. However, note that your answers vary only by a constant when we don't multiply by 2 (if we add in absolute values to their answer): 1/2(x^2 - 2x) + ln|x+2| + ln|x-1| + c1 = 1/2(x^2 - 2x + 1) + ln|x+2| + ln|x-1| + c2 1/2(x^2 - 2x) + ln|x+2| + ln|x-1| + c1 = 1/2(x^2 - 2x) + ln|x+2| + ln|x-1| + 1/2 + c2 c1 = 1/2 + c2 <-- They only vary by constant values Since c2 is just some random constant, we could also just combine 1/2 + c2 into one constant.

    • 2 years ago
  21. chrisplusian Group Title
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    ok that is kind of confusing. so my answer was right just without the minus two. Is there a way I can use wolfram to see if my answer is equivalent to the one they give me?

    • 2 years ago
  22. chrisplusian Group Title
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    I meant times two

    • 2 years ago
  23. Chlorophyll Group Title
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    = x²/2 - x + ln | x+2| + ln | x -1| + C

    • 2 years ago
  24. Chlorophyll Group Title
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    or x² - 2x + 2ln | x+2| + 2ln | x -1| + C

    • 2 years ago
  25. Chlorophyll Group Title
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    You have 4 separates terms to take derivative. The result looks neat without fraction !

    • 2 years ago
  26. Chlorophyll Group Title
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    *separated

    • 2 years ago
  27. chrisplusian Group Title
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    I am still somewhat confused I don't get how they are the same

    • 2 years ago
  28. Chlorophyll Group Title
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    The first one has fraction 1/2 If you don't want fraction form, you need to multiply all terms by 2!

    • 2 years ago
  29. Chlorophyll Group Title
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    = x²/2 - x + ln | x+2| + ln | x -1| + C x² - 2x + 2ln | x+2| + 2ln | x -1| + C

    • 2 years ago
  30. AccessDenied Group Title
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    How would you multiply by 2 without changing the value of the original integral? If you take the derivative of the indefinite integral multiplied by 2, you get the original function, multiplied by 2...

    • 2 years ago
  31. Chlorophyll Group Title
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    This is how you multiply: = 2 ( x²/2 - x + ln | x+2| + ln | x -1| ) + C = x² - 2x + 2ln | x+2| + 2ln | x -1| + C

    • 2 years ago
  32. chrisplusian Group Title
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    ok to start with I did that but then @AccessDenied said I could not. The basic problem I have is understanding how wolfram alpha's answer is the same as mine. @Chlorophyll could you scroll up to where i posted my answer as compared to my answer?

    • 2 years ago
  33. chrisplusian Group Title
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    I got (x^2)-2x+1 and I just cannot see how that is the same as (x-2)x

    • 2 years ago
  34. Chlorophyll Group Title
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    This is just plain straight forward formula from the textbook!

    • 2 years ago
  35. Chlorophyll Group Title
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    " I got (x^2)-2x+1 " Where do you get 1?

    • 2 years ago
  36. chrisplusian Group Title
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    sorry I must have missed the "plain straightforward" part of my calculus II textbook

    • 2 years ago
  37. chrisplusian Group Title
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    lol

    • 2 years ago
  38. Chlorophyll Group Title
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    Does it mean you understand your mistake?

    • 2 years ago
  39. AccessDenied Group Title
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    It is an indefinite integral, so your answer has one constant of variation added on. The difference between your answers is a constant, so it can be combined with your constant of variation since C + 1/2 is also a constant. :P

    • 2 years ago
  40. chrisplusian Group Title
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    I used the "x-1" as u then du=dx. so I then have the integral of u du. That evaluated becomes (u^2)/2. plug x-1 back in for u and then square it. that gives you (x^2)-2x+1 all divided by two

    • 2 years ago
  41. Chlorophyll Group Title
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    u = x -1 --> du = dx => ∫ du/ u = ln u = ln | x-1|

    • 2 years ago
  42. chrisplusian Group Title
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    it was just x-1 not 1/(x-1)

    • 2 years ago
  43. Chlorophyll Group Title
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    Oh, so they're 2 individual terms!

    • 2 years ago
  44. chrisplusian Group Title
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    can't you integrate them as I just described? I just ask cause that is where my mind went when I had to do it without any help

    • 2 years ago
  45. Chlorophyll Group Title
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    No, it doesn't work that way. We only apply substitution when we're unable to integral term by term: Suppose ∫ ( x - 1) dx Let u = x -1 => du = dx =>∫ udu = u²/2 = ( x-1)² /2 = ( x² - 2x + 1 ) /2 = x²/2 - x + 1/2

    • 2 years ago
  46. Chlorophyll Group Title
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    * to integrate

    • 2 years ago
  47. AccessDenied Group Title
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    = x²/2 - x + 1/2 [+ c] Let 1/2 + c = C (combine constants into a single constant of variation) = x²/2 - x + C

    • 2 years ago
  48. AccessDenied Group Title
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    The big thing about indefinite integration is that there is an additional constant of variation due to differentiation removing any constants and you are reversing the process. These answers only vary by constants, where differentiation removes both 1/2 and your constant of variation...

    • 2 years ago
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