Partial Fraction Decomposition when denominator can't be further factored

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SUMMARY

The discussion centers on the approach to partial fraction decomposition for the expression $$\frac{x^2}{x^2 + 9}$$ where the denominator cannot be factored further. The correct method involves rewriting the fraction as $$1 - \frac{9}{x^2 + 9}$$, allowing for integration rather than traditional partial fraction decomposition. This technique is a common strategy when the degrees of the numerator and denominator are equal.

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tmt1
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I have this fraction

$$x^2 / (x^2 + 9)$$

I'm not sure how to approach this problem since the denominator can't be further factored. What is the right approach for this type of problem?
 
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In order to do partial fractions we must have an expression that has a lower order in the numerator than in the denominator. In your case they are the same. So you would be looking for a form like [math]\frac{Ax + B}{x^2 + 9}[/math].

I would suggest looking at it like this:
[math]\frac{x^2}{x^2 + 9} = \frac{x^2 + 9 - 9}{x^2 + 9} = 1 - \frac{9}{x^2 + 9}[/math]
which is now in the desired form.

This is a fairly common trick.

-Dan
 
tmt said:
I have this fraction: $$x^2 / (x^2 + 9)$$

I'm not sure how to approach this problem since the denominator can't be further factored.
What is the right approach for this type of problem?
it depends on what you intend to do with it.

If you are trying to decompose it into Partial Fractrions, nothing can be done.

If you are trying to integrate it:

. . \int\frac{x^2}{x^2+9}dx \;=\;\int\frac{x^2+9 - 9}{x^2+9} dx

. . =\;\int\left(\frac{x^2+9}{x^2+9} - \frac{9}{x^2+9}\right)dx \;=\;\int\left(1 - \frac{9}{x^2+9}\right) dx \;\cdots\;\text{etc.}
 

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