Integration of Rational Functions with Partial Fractions

In summary, to integrate 1/(x^2-5x+6), the suggested method is to use partial fractions and rewrite the expression as A/(x-3) + B/(x-2). The values for A and B can be found by solving for them in the new equation. Once A and B are found, the integration would involve integrating 1/(x-2) and 1/(x-3) separately and setting the result equal to (t+c). Care must be taken with the signs of A and B in the final solution.
  • #1
alexis36
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Homework Statement


Integration of 1/(x^2-5x+6)


Homework Equations





The Attempt at a Solution


I know i cannot do ln|x^2-5x+6|
I've tried some form of substitution or intergration by parts, and they don't work.
Should I factor the bottom?
 
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  • #2
Have you tried partial fractions, as the title suggests? To be brief, you can rewrite [tex]\frac{1}{x^2-5x+6}[/tex] as [tex]\frac{a}{x-2}+\frac{b}{x-3}[/tex] where you must find a and b.
 
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  • #3
Okay.. I've solved the question up to a certain point. (I've decided that it's do-able without partial practions..)
however; now I am stuck on an integration:
i might find the integral of
1/(x^2-5x+6)
and now I am not sure if i use substituion, integration by parts, or bring the bottom up to the top of the fraction for this integration?
 
  • #4
alexis36 said:
I've solved the question up to a certain point. (I've decided that it's do-able without partial practions..)
Ok, show us then. Maybe we can help you from there.
 
  • #5
So partial fractions is the way to go I think..
I have
A/(x-3) + B/(x-2) = )A(x-2)+B(x-3))/(x-2)(x-3)
Ive solved for A and B and I got 1 for each of them.
Now I need to go and integrate 1/(x-2) and 1/(x-3) right? and set that equal to (t+c) and then solve the autonomous equation as i would any?
 
  • #6
Now I need to go and integrate 1/(x-2) and 1/(x-3) right?
Yes, you would integrate it from here.
 
  • #7
A=1 and B=1 doesn't work. Think about the signs again.
 

FAQ: Integration of Rational Functions with Partial Fractions

What are partial fractions in calculus?

Partial fractions in calculus is a method used to break down a complex rational expression into simpler fractions. This is done by expressing the numerator of the rational expression as a sum of simpler polynomials.

Why is partial fractions used in calculus?

Partial fractions are used in calculus to make it easier to integrate rational functions. By breaking down a complex rational expression into simpler fractions, it becomes easier to find the integral of each individual fraction.

How do you find the partial fraction decomposition of a rational expression?

To find the partial fraction decomposition of a rational expression, you need to follow a few steps. First, factor the denominator into irreducible polynomials. Then, for each distinct polynomial, set up a fraction with that polynomial as the denominator and an undetermined coefficient as the numerator. Finally, equate the coefficients of the like terms in the original rational expression and the partial fractions to solve for the undetermined coefficients.

Can you use partial fractions to solve indefinite integrals?

Yes, partial fractions can be used to solve indefinite integrals. By breaking down a complex rational expression into simpler fractions, you can then integrate each individual fraction using standard integration techniques.

Are there any limitations to using partial fractions in calculus?

There are a few limitations to using partial fractions in calculus. First, the rational expression must be proper, meaning the degree of the numerator is less than the degree of the denominator. Additionally, the denominator must be factorable into distinct linear or quadratic factors. If these conditions are not met, partial fractions cannot be used to solve the integral.

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