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abz_tech

  • 3 years ago

integration question

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  1. abz_tech
    • 3 years ago
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    \[\int\limits\limits_{0}^{1} \frac{ x+1 }{ x ^{2}+6x+5 }dx\]

  2. Mimi_x3
    • 3 years ago
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    \[x^2+ 6x + 5 = (x+3)^2 -(3)^2 + 5 = (x+3)^2 -9 +5 = (x+3)^2 -1 \] \[\int\limits \frac{x+1}{(x+3)^{2}-1} \]

  3. Mimi_x3
    • 3 years ago
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    Wait. \[\frac{x+1}{(x+5)(x+1)} = \frac{A}{(x+5)} +\frac{B}{(x+1)} \] Better now. :D

  4. abz_tech
    • 3 years ago
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    0_o

  5. Mimi_x3
    • 3 years ago
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    For A: \(x= -5\) \[\frac{(-5+1)}{(-5+1)} =\frac{-4}{-4}=1 \]

  6. Hero
    • 3 years ago
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    Aren't you forgetting something @Mimi_x3 ?

  7. Mimi_x3
    • 3 years ago
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    For B: \(x= -1\) => \(0\) What am I forgetting?

  8. Hero
    • 3 years ago
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    \[\frac{(x+1)}{(x+5)(x+1)} = \frac{1}{x+5}\]

  9. Mimi_x3
    • 3 years ago
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    There is something wrong.

  10. abz_tech
    • 3 years ago
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    isn't u's and du/dx supposed to come into play about now 0_o?

  11. Mimi_x3
    • 3 years ago
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    Oh crap! Sorry!

  12. abz_tech
    • 3 years ago
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    0_o

  13. Mimi_x3
    • 3 years ago
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    So, now \(u= x+5\)

  14. Hero
    • 3 years ago
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    \[\int\limits_{0}^{1}(x+5)^{-1}dx\]

  15. Mimi_x3
    • 3 years ago
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    I was right anyway :P just went the long way.

  16. Hero
    • 3 years ago
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    Ya, exactly. You're so silly :P

  17. Mimi_x3
    • 3 years ago
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    That doesn't work hero, u = x+5

  18. Hero
    • 3 years ago
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    I can make u = x + 5 if I want or not. It doesn't matter. It will still work.

  19. abz_tech
    • 3 years ago
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    how would u solve (x+5)^-1

  20. abz_tech
    • 3 years ago
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    \[\ln x+5?\]

  21. winterfez
    • 3 years ago
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    no i guess that does not work

  22. Mimi_x3
    • 3 years ago
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    \[\int\limits\frac{1}{x+5}dx \] \[u = x+5 => \frac{du}{dx} = 1 => du = dx \] \[\int\limits\frac{1}{u} du = \ln(x+5) + C\] @hero: writing in that form won't work.

  23. Hero
    • 3 years ago
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    \[\ln(|x+5|)\]

  24. Hero
    • 3 years ago
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    I don't need @mimi_x3 ( a kid) to tell me things, lol

  25. winterfez
    • 3 years ago
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    @hero right, is absolute value for ln

  26. Mimi_x3
    • 3 years ago
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    I am not a kid!! It's the truth it won't work in the form; of if you want to use that retarded formula.

  27. abz_tech
    • 3 years ago
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    hero how would u go about doing \[\int\limits_{0}^{1} (x+5)^{-1}\]

  28. Mimi_x3
    • 3 years ago
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    Using that form; wont get you anyway; and get lost.

  29. abz_tech
    • 3 years ago
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    fine mimi, can u tell me your way =/?

  30. Hero
    • 3 years ago
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    She already did. @winterfez, also provided some good steps.

  31. Mimi_x3
    • 3 years ago
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    i typed up the solution!!! scroll upp!!

  32. Hero
    • 3 years ago
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    I think I deserve a medal for pointing @mimi_x3 in the right direction.

  33. abz_tech
    • 3 years ago
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    then with the limits would i be ln 6 - ln 5?

  34. Mimi_x3
    • 3 years ago
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    Well, first I was not wrong,I just went the long way. I would have gotten the right anwer without your input. Secondly, can you please show us how you would integrate \[\int\limits(x+5)^{-1} dx\]

  35. Hero
    • 3 years ago
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    That's top secret info.

  36. Mimi_x3
    • 3 years ago
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    Lol, because it's impossible :)

  37. Mimi_x3
    • 3 years ago
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    Me want a medal as well!

  38. Hero
    • 3 years ago
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    Nothing is impossible

  39. Mimi_x3
    • 3 years ago
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    :D

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