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chrisplusian

Quick integration question

  • one year ago
  • one year ago

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

    • one year ago
  2. chrisplusian
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    is this the same as:

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

    • one year ago
  4. AccessDenied
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    Yes, that would be correct.

    • one year ago
  5. chrisplusian
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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

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

    • one year ago
  7. chrisplusian
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    are these answers the same:

    • one year ago
  8. chrisplusian
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    \[x^2 -2x +1 +2\ln \left| x+2 \right| +2\ln \left| x-1 \right|\]

    • one year ago
  9. chrisplusian
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    and

    • one year ago
  10. chrisplusian
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    \[\frac{ 1 }{ 2 }(x-2)x +\ln (x-1) +\ln (x+2)\]

    • one year ago
  11. chrisplusian
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    ??

    • one year ago
  12. chrisplusian
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    The first one is my answer the second is wolfram alpha

    • one year ago
  13. AccessDenied
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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.

    • one year ago
  14. chrisplusian
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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?

    • one year ago
  15. AccessDenied
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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.

    • one year ago
  16. chrisplusian
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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.

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

    • one year ago
  18. chrisplusian
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    and multiply through by two to clear the fraction?

    • one year ago
  19. AccessDenied
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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.

    • one year ago
  20. AccessDenied
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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.

    • one year ago
  21. chrisplusian
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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?

    • one year ago
  22. chrisplusian
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    I meant times two

    • one year ago
  23. Chlorophyll
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    = x²/2 - x + ln | x+2| + ln | x -1| + C

    • one year ago
  24. Chlorophyll
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    or x² - 2x + 2ln | x+2| + 2ln | x -1| + C

    • one year ago
  25. Chlorophyll
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    You have 4 separates terms to take derivative. The result looks neat without fraction !

    • one year ago
  26. Chlorophyll
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    *separated

    • one year ago
  27. chrisplusian
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    I am still somewhat confused I don't get how they are the same

    • one year ago
  28. Chlorophyll
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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!

    • one year ago
  29. Chlorophyll
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    = x²/2 - x + ln | x+2| + ln | x -1| + C x² - 2x + 2ln | x+2| + 2ln | x -1| + C

    • one year ago
  30. AccessDenied
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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...

    • one year ago
  31. Chlorophyll
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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

    • one year ago
  32. chrisplusian
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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?

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

    • one year ago
  34. Chlorophyll
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    This is just plain straight forward formula from the textbook!

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

    • one year ago
  36. chrisplusian
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    sorry I must have missed the "plain straightforward" part of my calculus II textbook

    • one year ago
  37. chrisplusian
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    lol

    • one year ago
  38. Chlorophyll
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    Does it mean you understand your mistake?

    • one year ago
  39. AccessDenied
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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

    • one year ago
  40. chrisplusian
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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

    • one year ago
  41. Chlorophyll
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    u = x -1 --> du = dx => ∫ du/ u = ln u = ln | x-1|

    • one year ago
  42. chrisplusian
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    it was just x-1 not 1/(x-1)

    • one year ago
  43. Chlorophyll
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    Oh, so they're 2 individual terms!

    • one year ago
  44. chrisplusian
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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

    • one year ago
  45. Chlorophyll
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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

    • one year ago
  46. Chlorophyll
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    * to integrate

    • one year ago
  47. AccessDenied
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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

    • one year ago
  48. AccessDenied
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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...

    • one year ago
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