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monroe17

integral of (3x+5)/(5x^2-4x+1)dx

  • one year ago
  • one year ago

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  1. wio
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    It's a rational polynomial, have you tried factoring it?

    • one year ago
  2. monroe17
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    let me do it real quick

    • one year ago
  3. monroe17
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    it cant be factored..

    • one year ago
  4. wio
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    imaginary roots?

    • one year ago
  5. monroe17
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    huh?

    • one year ago
  6. AnElephant
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    imaginary roots wont help you in this i think... hold on... (props for a challenging question)

    • one year ago
  7. wio
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    You said it can't be factored... what are the roots of the denominator?

    • one year ago
  8. monroe17
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    2/5-i/5, 2/5+i/5?

    • one year ago
  9. wio
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    rational functions like this often don't have elementary anti derivatives.

    • one year ago
  10. wio
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    Is this a homework problem?

    • one year ago
  11. monroe17
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    it's a practice problem for my exam this coming thursday. I'm trying to study and get familiar with the problems.

    • one year ago
  12. AnElephant
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    ok got it, in this type of question you need to rewrite it so that at least one of the parts of the numerator is equal to the derivative of the denominator. So we may rewrite it like so: 31/5 integral(1/5x^2-4x+1) +3/10 integral((10x-4)/(5x^2-4x+1)) Use u substitution for integral((10x-4)/(5x^2-4x+1)) u= the denominator and du = the numerator So now you get 3/10 1/u du (which is also the same as 3/10 ln(u) Now substitute x back into u Now, rewrite the other part of the integral (complete the square) to find an inverse tan function hidden there. It simplifies to the equation 31/5 (arctan(2-5x) so the final equation is : 3/10 ln(5x^2-4x+1) + 31/5 (arctan(2-5x) +c (do NOT forget the +C after all this work lol)

    • one year ago
  13. AnElephant
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    if you want a more insightful thought process of a certain part of a problem, please ask (i was kinda just in a hurry to get it done)

    • one year ago
  14. monroe17
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    woah woah, explain how you rewrote it.. i dont get that..

    • one year ago
  15. AnElephant
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    Okay, so, we know the derivative of the denominator is 10x-4 right? so we just have to do a lot of algebra to get at least one part of the numerator to look like that (its mostly a lucky guess in my case) because i had in the back of my mind that a problem as complex as this needs to use u substitiution

    • one year ago
  16. monroe17
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    so, i have to get the numerator to be 10x-4? (which is the derivative of the denom)

    • one year ago
  17. AnElephant
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    3x+5 = 31/5 + 3/10 (10x-4) ; (the 2 numerators of the integrals before i factored out constants

    • one year ago
  18. AnElephant
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    Yes, in a problem where there is a denominator, you need to first think of substitution and how you can get the derivative of the denominator into the numerator of the function

    • one year ago
  19. monroe17
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    I still don't get how in the worl you get 31/5+3/10(10x-4)... out of 3x+5

    • one year ago
  20. AnElephant
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    ok, so, we need to get 10x-4 out of this so : 3x+5 = 30x+50, then subtract 3(10x-4) to get a remainder of 62 (this will go in the 2nd integral)

    • one year ago
  21. AnElephant
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    keep in mind, this is all over a denominator of 10 (which we inserted into the equation when we multiplied 3x+5 by 10)

    • one year ago
  22. monroe17
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    so multiply by ten and subtract 3(10x-5)?

    • one year ago
  23. AnElephant
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    yeah, and hope you get lucky later on ; )

    • one year ago
  24. monroe17
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    how do you subtract 3(10x-5) from 30x+50? lol! im so sorry... i just really dont understand

    • one year ago
  25. AnElephant
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    no problem, ive been told i make a horrible teacher XD (and it's 3(10x-4) So, 30x+50 - 3(10x-4) = 30x+50 - 30x + 12 which makes 62 (the constant in the 2nd integral) then you simply divide by 10 to get 31/5 which is what i factored out of the integral

    • one year ago
  26. monroe17
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    ahhh, got it,. what you mean constant in the 2nd integral,. ?

    • one year ago
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