ggrree
  • ggrree
http://gyazo.com/7919fbc4610aa7647d33100ee01aec29
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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ggrree
  • ggrree
I was able to separate the integrand into partial fractions but then I get stuck.
myininaya
  • myininaya
Is that a type-o?
myininaya
  • myininaya
\[\int\limits_{0}^{9}\frac{dx}{x^2-6+5}\]

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myininaya
  • myininaya
Is it really suppose to be x^2-6+5 and not x^2-6x+5?
myininaya
  • myininaya
\[\frac{1}{x^2-1} \text{ since } -6+5=-1\] \[\frac{1}{(x-1)(x+1)}=\frac{A}{x-1}+\frac{B}{x+1}\]
myininaya
  • myininaya
\[\frac{1}{(x-1)(x+1)}=\frac{A(x+1)+B(x-1)}{(x-1)(x+1)}\] => 1=x(A+B)+(A-B) A+B=0 ; A-B=1
myininaya
  • myininaya
Can you do the test now?
myininaya
  • myininaya
And by the way this is an improper integral meaning the function is not continuous on [0,9]
myininaya
  • myininaya
You will have to break up the integral
myininaya
  • myininaya
\[\int\limits_{0}^{1}\frac{dx}{x^2-1} +\int\limits_{1}^{9}\frac{dx}{x^2-1}\]
myininaya
  • myininaya
See if both parts converge if so then take the sum in you are done if not then the area/net area is divergent
myininaya
  • myininaya
lol I meant rest above not test
myininaya
  • myininaya
@ggrree any questions?
apoorvk
  • apoorvk
i think we have a standard formula as well for integrals of this type rather than use partial fractions every time: \[\int\limits(1/(x^2-a^2))dx =(1/2a)\ln((x-a)/(x+a)) + k\]
apoorvk
  • apoorvk
|dw:1332635709657:dw|
apoorvk
  • apoorvk
@ggrree
ggrree
  • ggrree
Yeah, I think I got it. I think... I think it diverges. is that right?
ggrree
  • ggrree
thanks guys!
myininaya
  • myininaya
you are right! @ggrree does not converge :)

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