# Partial Fractions Decomposition

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1 Partial Fractions Decomposition Dr. Philippe B. Laval Kennesaw State University August 6, 008 Abstract This handout describes partial fractions decomposition and how it can be used when integrating rational functions. Partial Fraction Decomposition. Introduction This handout describes a method to rewrite a fraction we do not know how to integrate into simpler (partial) fractions we know how to integrate. This method works for rational functions, that is functions which can be written as the quotient of two polynomial functions. For the remaining part of this document, we will assume that we have a rational function p () in which degree of p () < degree of q (). If this is not q () the case, we can always perform long division. For eample, if we were given 3 the fraction. We would perform long division to obtain 3 = + 3 = + We then would apply partial fraction decomposition to. In this class, we will use partial fraction decomposition as an integration technique. The ultimate goal is to decompose a fraction so we can integrate it. In this document, we will focus on the decomposition. Keep in mind why we are doing this decomposition. How we perform partial fraction decomposition depends of the denominator of the fraction. We consider several cases.

2 . Case : q() is a product of distinct linear factors. Let us assume that q () is a product of n distinct linear factors that is Then, q () = (a + b ) (a + b ) ::: (a n + b n ) p () q () = A a + b + A a + b + ::: + a n + b n Finding the decomposition amounts to nding the coe cients A,..., A n. This can be done two di erent ways. We illustrate this in the eamples below. Remark p () does not play a role in the way the decomposition is written. Eample Find the decomposition for + 3 We begin by factoring the denominator. We obtain ( + 3) ( ). According to the decomposition we wrote above, we have: ( + 3) ( ) = A B We need to nd A and B. Multiplying each side by the denominator of the fraction on the left and simplifying, we obtain: = A ( ) + B ( + 3) = A A + B + 3B = (A + B) A + 3B A n Two polynomials are equal if their corresponding coe cients are equal. gives us the following system: A + B = This A + 3B = 0 8 >< A = 3 The solution of this system is: 4 >: B = Thus, we have = There is an easier way which works in this case. We show it on the net eample. Eample 3 Find the decomposition for ( ) ( + 5) ( 3) The denominator is already factored. The decomposition is: ( ) ( + 5) ( 3) = A + B C 3

3 We need to nd A, B, and C. We begin the same way, we multiply each side by the denominator of the fraction on the left, and simplify. We obtain: = A ( + 5) ( 3) + B ( ) ( 3) + C ( ) ( + 5) Then we notice that since the above equality is true for every, it will be true for speci c values of. We select values for which will make all but one of the coe cients go away. We will then be able to solve for that coe cient. More precisely, When =, we obtain: 0 = A (6) ( ) A = 0 When = 5, we obtain 30 = B ( 6) ( 8) B = B = 5 8 When = 3, we obtain 6 = C () (8) Therefore, the decomposition is: C = 3 8 ( ) ( + 5) ( 3) = Case : q() is a product of linear factors, some being repeated The factors which are not repeated will be decomposed as above. Suppose that q () also contains (a + b) m that is a + b is repeated m times. The decomposition for this factor will be Eample 4 Find a decomposition for The decomposition is: A a + b + A (a + b) + ::: + A m (a + b) m ( ) ( + ) 3 ( ) ( + ) 3 = A + B + + C ( + ) + D ( + ) 3 3

4 Then, we proceed as before. We multiply each side by the denominator of the fraction on the left and simplify. We obtain: = A ( + ) 3 + B ( ) ( + ) + C ( ) ( + ) + D ( ) We need to nd A, B, C, and D. We can use either of the methods described in the rst case. If =, we get = A A = If =, we get = 3D D = 3 We still have to nd B and C. For this, we choose two more values for and write the corresponding system. We now know A and D, so we can use the value we found for them. If =, we get If = 0, we get = 8A 4B C D = 6 4B + C = 6 4B + C = 34 4B C = A B C D = B + C = B + C = 38 B C Thus, we need to solve 8 >< >: 4B + C = 34 B + C = 38 4

5 The solution is : B = ; C = 7 9. Putting all this together, we get ( ) ( + ) 3 = ( ) ( + ) ( + ) + 3 ( + ) 3.4 Case 3: q() is a product of distinct irreducible quadratic factors. Recall that a term is called irreducible if it cannot be factored any further. Thus + + is irreducible, so is +. Be careful, ( + ) is not considered a quadratic term. You must think of it as a linear term appearing twice. We can generalize what we did in the previous two cases as follows. Instead of thinking of linear factors, think that when we write the decomposition, the degree of the term in the numerator is less than the degree of the term in the denominator. When we had linear factors in the denominator, it meant that we had to have terms of degree 0 in the numerator, that is we had constant terms. If the denominator contains irreducible quadratic factors, then the numerator will contain linear terms. We look at an eample to see how the decomposition is written. Eample 5 Decompose ( + ) ( + ) If we apply a method similar to that of case, the decomposition will contain two fractions, one for each irreducible factor. However, since the degree of the denominator is now, the degree of the numerator will be, that is we will have linear terms. Recall that a linear term is of the form a + b. Thus, ( + ) ( + ) = A + B + + C + D + To solve, we proceed as above. First, we multiply each side by the denominator of the fraction on the left, and simplify. We obtain: = (A + B) + + (C + D) + We then pick 4 di erent values for to get a system of 4 equations, which we solve. The answer is: ( + ) ( + ) = Case 4: q() is a product of irreducible quadratic factors, some being repeated This is similar to case, with linear terms in the numerator and quadratic terms in the denominator. 5

6 Eample 6 Write the decomposition for ( + + ) 3 ( + + ) 3 = A + B C + D ( + + ) + E + F ( + + ) 3 We would nd the coe cients as above..6 General Case: q() is a miture of the above Eample 7 Decompose ( ) ( + + ) 3 ( ) ( + + ) 3 = A + B ( ) The answer is + C + D E + F ( + + ) + G + H ( + + ) 3 ( ) ( + + ) 3 = ( ) ( ) ( + + ) 3 ( + + ) 3.7 Application The idea behind this decomposition is that once the fraction is decomposed, we can integrate it. Eample 8 Find + 3 d We found earlier that + 3 = Therefore, + 3 d = d d Using substitution Eample 9 Find ( + ) ( + ) d = 3 4 ln j + 3j + ln j j 4 We saw earlier as an eample that the partial fraction decomposition of was ( + ) ( + ) = ( + ) ( + )

7 Therefore ( + ) ( + ) d = + d + d We can do the rst integral by substitution. If we let u = +, then du = d and therefore + d = du u = ln juj = ln + The second integral is done in a similar way. We obtain + d = ln + It follows that ( + ) ( + ) d = ln + ln + = ln + + s + = ln + + C Eample 0 Find R 3 d The function we are integrating is a rational function. However, the degree of the numerator is greater than or equal to the degree of the denominator. So, the rst step is to perform long division. We did this at the beginning of this document and found that It follows that 3 = + 3 d = d + d () We can do the rst integral. The second, is the integral of a rational function. To be able to evaluate it, we rst decompose into partial fractions. = ( ) ( + ) = A + B + 7

8 We need to nd A and B. We do it using the techniques described above. First, we multiply each side by the denominator of the fraction on the left to obtain = A ( + ) + B ( ) When =, we get = A or A =. When =, we get = B or B =. Therefore, = + + If we replace what we just found in equation, we obtain 3 d = d + d + d + These are integrals we can handle. 3 d = + ln j j + ln j + j.8 Problems Do the problems below: 4. Find 4 ( ) ( + ) d. Find + d + 3. Find d d 4. Find where a 6= 0 a Find 0 ( + ) d d 6. Find ( ) ( + 4) = + ln 7. Do # 5, 6, 7, 9,, 5, on page 405 8

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LINEAR INEQUALITIES When we use the equal sign in an equation we are stating that both sides of the equation are equal to each other. In an inequality, we are stating that both sides of the equation are

### Lesson 9: Radicals and Conjugates

Student Outcomes Students understand that the sum of two square roots (or two cube roots) is not equal to the square root (or cube root) of their sum. Students convert expressions to simplest radical form.

### UNIT 1: ANALYTICAL METHODS FOR ENGINEERS

UNIT : NLYTICL METHODS FOR ENGINEERS Unit code: /0/0 QCF Level: Credit value: OUTCOME TUTORIL LGEBRIC METHODS Unit content Be able to analyse and model engineering situations and solve problems using algebraic

### Solving Equations of Degree 1 (Linear Equations):

Solving Equations of Degree 1 (Linear Equations): Definition: The equation a b0 is called an equation of degree 1. Eample: Solve for : 0. 0, Or. Dividing both sides by : giving. Easy! Check the answer!