Definite Integration - but how to integrate?

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Homework Help Overview

The discussion revolves around the integration of a rational function, specifically involving the expression \( \frac{1}{x^2 - 9} \) and its definite integral from 4 to 5. Participants are exploring methods to approach the integration, including the use of partial fractions and potential trigonometric substitutions.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of partial fractions to simplify the integral and question the correct application of limits after resolving the fractions. There is also mention of potential trigonometric identities and the confusion surrounding the interpretation of \( dx \) in the context of integration.

Discussion Status

The discussion is active, with participants providing suggestions on how to approach the problem, such as using partial fractions and checking calculations. There is an ongoing exploration of the correct values to substitute and how to simplify the resulting expressions, but no consensus has been reached on the final approach or solution.

Contextual Notes

Participants are also seeking guidance on using equation editing software for formatting their mathematical expressions, indicating a need for clarity in communication within the thread.

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Firstly apologies for the calculation, where can I use equation editing software/fonts?

Show that :
dx = 1/6ln(7/4) where the limits are : b = 5 and a = 4
x^2 -9




I know that if the denominator was x^2 + 9 I could use 1/a tan-1 x/a + C . But do not know how to attempt this - does it involve the substitution of a trig identity, if so which one...? Thank you.
 
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Partial fractions.

[tex]\frac{1}{x^2-9} = \frac{A}{x+3} + \frac{B}{x-3}[/tex]. Find A and B, then this integral becomes easy.
 
Ok right so whenever I see dx as the numerator I presume it's 1? I had resolved this into partial fractions on a previous part of the question, so..
. 1 .=
x^2 -9

. 1 . - . 1 .
6(x-3) .6(x+3)

And then with the limits do I substitute values into x of the partial fractions?

(1/6 ln(2) - 1/6 ln (8)) - (1/6 ln (1) - 1/6ln(7))
If this is right how do I get to the answer given, or have i missed something completely?
Many thanks,

any assistance with the equation writing font/software?
 
With your posts, click "Go Advanced" and just above the text input box, starting with the Bold button, at the end of that row there's a Sigma letter. Click that and play around with it and the preview post =]

You don't presume its 1, you just realize that [tex]\frac{dx}{f(x)} = \frac{1}{f(x)} dx[/tex]. I'm not sure about your values of x, cross multiply them back and see if it comes back to the original expression. Then make sure you plug your values in correctly, both both integrals. Then you might be able to simplify it with some log identities.
 

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