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Mathematics
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I was able to separate the integrand into partial fractions but then I get stuck.
Is that a type-o?
\[\int\limits_{0}^{9}\frac{dx}{x^2-6+5}\]

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Other answers:

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

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