MHB Partial Fraction Decomposition when denominator can't be further factored

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To approach the fraction x^2 / (x^2 + 9), recognize that the numerator and denominator have the same degree, which complicates partial fraction decomposition. Instead, rewrite the expression as 1 - (9 / (x^2 + 9)), allowing for easier integration. This method is a common technique for handling such fractions. If the goal is integration, this transformation simplifies the process significantly. Understanding the relationship between the degrees of the numerator and denominator is crucial for determining the appropriate approach.
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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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