INTEGRATION BY PARTIAL FRACTIONS        

Phoebe 

 

Integration by partial fractions is a useful method to find the antiderivative of a function with a factorable (or factored) denominator. The approach is a simple yet practical one. To make the integration simpler, we will rewrite the integrand as the sum of fractions and integrate them separately. The same rule that requires a common denominator to add fractions will be used to integrate.

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WHAT TO LOOK OUT FOR

This form of integration is the easiest one to recognize.

The most recognizable sign is that the integrand is a fraction with factorable (or factored) denominator.

Here are some examples of what you might find.

 

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HOW TO APPROACH IT

(carefully- factoring can be a ten-headed beast for some of us)

 

Let's take the first example. 

Remember it is

 

For now ignore the integral sign and split the denominator into two fractions that equal the numerator with constant numerators A and B. These constants will always exist for the rational function p(x)/q(x) when p(x) and q(x) are both polynomials AND the degree of p(x) is less than the degree of q(x).

http://www.math.ucdavis.edu/~kouba/CalcTwoDIRECTORY/partialfracdirectory/PartialFrac.html 

      

Next cross multiply the left side of the equal sign so that each fraction has the same denominator. 

     

     Since all three fractions have the same denominator, we can simplify this to

     

Now substitute different values for x so that each fraction equals zero. 

     let x=-5

      

      therefore

     

     next, let x=2

      

      therefore 

Now incorporate your integral sign and the values you have just found, integrate, and manipulate. 

       

     

     

     

     

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STEPS IN ACTION 

If the numerator also has a variable, just follow the same steps as above. 

Let's take the second example. Remember it's  .

The denominator of this integrand can be factored, so the integral we will work with looks like .

Split

     

Cross multiply

     

Substitute

     let x=1

     

      therefore

     next, let x=-3/2

     

      therefore 

Incorporate, integrate, and manipulate

      

     

     

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WARNING

This can get a bit tricky if you don't use integration by partial fractions when necessary. For example, on one of Alexis's recent tests she tried to use integration by substitution to evaluate . Needless to say it did not work out very well.

 

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APPLETS FOR EXTRA EXPLANATIONS

http://archives.math.utk.edu/visual.calculus/4/partial_fractions.2/index.html 

(This link solves for the values of A and B as a system of equations, but the general idea and order of steps are the same as discussed above)

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EXAMPLE PROBLEMS

1. Show Alexis the correct way to evaluate .

2. Evaluate .

3. Evaluate .

4. Which of the following is equal to ? 

     A.

     B.

     C.

     D.

5. Which of the following would require the use of integration by partial fractions? Choose more than one if necessary.

     A.

     B.

     C.

     D.

6. Evaluate .

     A. 

     B. 

     C. 

     D. 

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ANSWERS 

1.  

2.

3.

4. C

5. A, C- This one is a little tricky. D is not correct because the antiderivative is arctan the squared term. B is also not correct because it requires you to complete the square, which will result in arctan of the squared term again.

Here is a full explanation of B. Number 3 on the applet link will show an explanation of C.

6. A

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HELPFUL EXTERNAL LINKS

http://www.math.ucdavis.edu/~kouba/CalcTwoDIRECTORY/partialfracdirectory/PartialFrac.html

http://en.wikipedia.org/wiki/Partial_fractions_in_integration 

http://www.karlscalculus.org/calc11_5.html