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  1. ggrree
    • 4 years ago
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    I was able to separate the integrand into partial fractions but then I get stuck.

  2. myininaya
    • 4 years ago
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    Is that a type-o?

  3. myininaya
    • 4 years ago
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    \[\int\limits_{0}^{9}\frac{dx}{x^2-6+5}\]

  4. myininaya
    • 4 years ago
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    Is it really suppose to be x^2-6+5 and not x^2-6x+5?

  5. myininaya
    • 4 years ago
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    \[\frac{1}{x^2-1} \text{ since } -6+5=-1\] \[\frac{1}{(x-1)(x+1)}=\frac{A}{x-1}+\frac{B}{x+1}\]

  6. myininaya
    • 4 years ago
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    \[\frac{1}{(x-1)(x+1)}=\frac{A(x+1)+B(x-1)}{(x-1)(x+1)}\] => 1=x(A+B)+(A-B) A+B=0 ; A-B=1

  7. myininaya
    • 4 years ago
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    Can you do the test now?

  8. myininaya
    • 4 years ago
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    And by the way this is an improper integral meaning the function is not continuous on [0,9]

  9. myininaya
    • 4 years ago
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    You will have to break up the integral

  10. myininaya
    • 4 years ago
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    \[\int\limits_{0}^{1}\frac{dx}{x^2-1} +\int\limits_{1}^{9}\frac{dx}{x^2-1}\]

  11. myininaya
    • 4 years ago
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    See if both parts converge if so then take the sum in you are done if not then the area/net area is divergent

  12. myininaya
    • 4 years ago
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    lol I meant rest above not test

  13. myininaya
    • 4 years ago
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    @ggrree any questions?

  14. apoorvk
    • 4 years ago
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    i think we have a standard formula as well for integrals of this type rather than use partial fractions every time: \[\int\limits(1/(x^2-a^2))dx =(1/2a)\ln((x-a)/(x+a)) + k\]

  15. apoorvk
    • 4 years ago
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    |dw:1332635709657:dw|

  16. apoorvk
    • 4 years ago
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    @ggrree

  17. ggrree
    • 4 years ago
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    Yeah, I think I got it. I think... I think it diverges. is that right?

  18. ggrree
    • 4 years ago
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    thanks guys!

  19. myininaya
    • 4 years ago
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    you are right! @ggrree does not converge :)

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