# Integrating algebraic fractions

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1 Integrating algebraic fractions Sometimes the integral of an algebraic fraction can be found by first epressing the algebraic fraction as the sum of its partial fractions. In this unit we will illustrate this idea. We will see that it is also necessary to draw upon a wide variety of other techniques such as completing the square, integration by substitution, using standard forms, and so on. In order to master the techniques eplained here it is vital that you undertake plenty of practice eercises so that they become second nature. After reading this tet, and/or viewing the video tutorial on this topic, you should be able to: integrate algebraic fractions by first epressing them in partial fractions integrate algebraic fractions by using a variety of other techniques Contents. Introduction. Some preliminary results. Algebraic fractions with two linear factors 4. Algebraic fractions with a repeated linear factor 5 5. Dealing with improper fractions 6 6. Algebraic fraction with an irreducible quadratic factor 7 c mathcentre August 8, 004

2 . Introduction In this section we are going to look at how we can integrate some algebraic fractions. We will be using partial fractions to rewrite the integrand as the sum of simpler fractions which can then be integrated separately. We will also need to call upon a wide variety of other techniques including completing the square, integration by substitution, integration using standard results and so on. Eamples of the sorts of algebraic fractions we will be integrating are ( )( + ), + +, ( ) ( + ) and 4 Whilst superficially they may look similar, there are important differences. For eample, the denominator of the first contains two linear factors. The second has an irreducible quadratic factor (i.e. it will not factorise). The third contains a factor which is repeated. The fourth is an eample of an improper fraction because the degree of the numerator is greater than the degree of the denominator. All of these factors are important in selecting the appropriate way to proceed.. Some preliminary results To understand the eamples which follow you will need to use various techniques which you should have met before. We summarise them briefly here, but you should refer to other relevant material if you need to revise the details. Partial fractions a linear factor, a + b in the denominator gives rise to a partial fraction of the form A a + b. repeated linear factors, (a + b) give rise to partial fractions of the form A a + b + B (a + b) a quadratic factor a + b + c gives rise to a partial fraction of the form Integration - standard results f () d ln f() + c f() e.g. a + d a tan a + c Integration - substitution ( ) du To find ( ) d, substitute u, du d to give d ( ) d u du u du A + B a + b + c d ln c u + c + c c mathcentre August 8, 004

3 . Algebraic fractions with two linear factors In this section we will consider how to integrate an algebraic fraction which has the form of a proper fraction with two linear factors in the denominator. Eample Suppose we want to find ( )( + ) d Note that the integrand is a proper fraction (because the degree of the numerator is less than the degree of the denominator), and also that the denominator has two, distinct, linear factors. Therefore the appropriate form for its partial fractions is ( )( + ) A ( ) + B ( + ) where A and B are constants which we shall determine shortly. We add the two terms on the right-hand side together again using a common denominator. ( )( + ) A ( ) + B ( + ) A( + ) + B( ) ( )( + ) Because the fraction on the left is equal to that on the right for all values of, and because their denominators are equal, then their numerators too must be equal. So, from just the numerators, A( + ) + B( ) () We now proceed to find the values of the constants A and B. We can do this in one of two ways, or by miing the two ways. The first way is to substitute particular values for. The second way is to separately equate coefficients of constant terms, linear terms, quadratic terms etc. Both of these ways will be illustrated now. c mathcentre August 8, 004

4 Substitution of particular values for Because epression () is true for all values of we can substitute any value we choose for. In particular, if we let the first term on the right becomes zero, and everything looks simpler: 0 + B( ( )) from which 5B and so B 5 Similarly, substituting in epression () makes the second term zero: from which 5A A 5 Thus the partial fractions are ( )( + ) 5( ) 5( + ) Both of the terms on the right can be integrated: 5( ) 5( + ) d 5 d 5 + d 5 ln ln + + c 5 By using partial fractions we have broken down the original integral into two separate integrals which we can then evaluate. Equating coefficients A second technique for finding A and B is to equate the coefficients of equivalent terms on each side. First of all we epand the brackets in Equation () and collect together like terms: A + B + (A B) Equating constant terms on each side of this epression gives Equating the coefficients of on each side: 0 A + B () A B () These are two simultaneous equations we can solve to find A and B. Multiplying Equation () by gives Now, adding () and (4) eliminates the B s to give A B (4) 5A from which A 5. Also, from (), B A 5 just as we obtained using the 5 method of substituting specific values for. Often you will find that a combination of both techniques is efficient. c mathcentre August 8, 004 4

5 Eample Suppose we want to evaluate ( + ) d. Note again that the integrand is a proper fraction and also that the denominator has two, distinct, linear factors. Therefore the appropriate form for its partial fractions is ( + ) A + B ( + ) where A and B are constants we need to find. We add the two terms on the right-hand side together again using a common denominator. ( + ) A + B ( + ) A( + ) + B ( + ) Because the fraction on the left is equal to that on the right for all values of, and because their denominators are equal, then their numerators too must be equal. So, from just the numerators, A( + ) + B If we substitute 0 we can immediately find A: so that A. If we substitute we find B: A(0 + ) + B(0) A( + ) + B( ) so that B. Then ( + ) d + d Eercises [ ln ln + ] ( ln ln ) ( ln ln) 6 ln ln ln 6 ln Find each of the following integrals by epressing the integrand in partial fractions. + (a) d (b) d (c) ( + )( + ) ( + )( 4) ( )( + 7) d 5 c mathcentre August 8, 004

6 4. Algebraic fractions with a repeated linear factor When the denominator contains a repeated linear factor care must be taken to use the correct form of partial fractions as illustrated in the following eample. Eample Find ( ) ( + ) d. In this Eample there is a repeated factor in the denominator. This is because the factor appears twice, as in ( ). We write ( ) ( + ) A + B ( ) + C + A( )( + ) + B( + ) + C( ) ( ) ( + ) As before, the fractions on the left and the right are equal for all values of. Their denominators are equal and so we can equate the numerators: A( )( + ) + B( + ) + C( ) () Substituting in Equation () gives B, from which B. Substituting gives 4C from which C 4. Knowing B and C, substitution of any other value for will give the value of A. For eample, if we let 0 we find A + B + C and so A from which A. Alternatively, we could have epanded the right-hand side of Equation 4 (), collected like terms together and equated coefficients. This would have yielded the same values for A, B and C. The integral becomes ( ) ( + ) d 4( ) + ( ) + 4( + ) d 4 ln ( ) + ln + + c 4 Using the laws of logarithms this can be written in the following alternative form if required. 4 ln + ( ) + c Eercises. Integrate each of the following by epressing the integrand in partial fractions. + + (a) d (b) d (c) ( + ) ( ) ( + ) ( + ) ( 7) d. c mathcentre August 8, 004 6

7 5. Dealing with improper fractions When the degree of the numerator is greater than or equal to the degree of the denominator the fraction is said to be improper. In such cases it is first necessary to carry out long division as illustrated in the net Eample. Eample Find 4 d. The degree of the numerator is greater than the degree of the numerator. This fraction is therefore improper. We can divide the denominator into the numerator using long division of fractions. ) So Note that the denominator of the second term on the right hand side is the difference of two squares and can be factorised as 4 ( )( + ). So, ( )( + ) A + B + A( + ) + B( ) ( )( + ) As before, the fractions on the left and on the right are equal for all values of. Their denominators are the same, and so too must be their numerators. So we equate the numerators to give 4 A( + ) + B( ) Choosing we find 8 4A so that A. Choosing gives 8 4B so that B. So with these values of A and B the integral becomes 4 d d Eercises + ln + ln + + c. Use long division and partial fractions to find the following integrals (a) d (b) d (c) d (d) d 7 c mathcentre August 8, 004

8 6. Algebraic fractions with an irreducible quadratic factor When the denominator contains a quadratic, a +b+c, which will not factorise into two linear factors (said to be an irreducible quadratic factor) the appropriate form of partial fractions is A + B a + b + c Suppose the value of A turns out to be zero, then we have an integrand of the form constant a + b + c. A term like this can be integrated by completing the square as in the following eample. Eample Suppose we wish to find + + d. We start by completing the square in the denominator to give + + d ( + d ) + 4 If we now substitute u + we obtain u + du. There is a standard result which we 4 quote that + a d a tan a + c This enables us to complete this eample, with a /, and obtain terms of the original variable,, we have + + d + tan + c tan u. In / / On the other hand, if A is non-zero it may turn out that the numerator is the derivative of the denominator, or can be easily made so. In such cases, the standard result f () d ln f() + c f() can be used. Consider the following eample. Eample Suppose we wish to find d. Here the derivative of + + is + and so immediately we can write down the answer: d ln c Finally, we may have values of A and B such that the numerator is not the derivative of the denominator. In such a case we have to adjust the numerator to make it so, and then compensate by including another term as illustrated in the following eample. c mathcentre August 8, 004 8

9 Eample Suppose we wish to find + + d. f () Again we shall use the result that d ln f() + c but unfortunately the numerator f() is not quite equal to the derivative of the denominator. We would like the denominator to be +. We can adjust the integrand by writing it as follows: + + ( ) + + This has the effect of introducing the required in the numerator. To obtain the require +, we write ( ) ( ) + ( ) Each of these terms can be integrated using the results of the two previous eamples and we obtain + + d ln tan + c So you see that you need a combination of a variety of techniques together with some ingenuity and skill. Practice is essential so that you eperience a wide variety of such problems. Eample Find ( + ) d. As before we epress the integrand in partial fractions noting in particular the appropriate form to use when dealing with an irreducible quadratic factor. ( + ) A + B + C + A( + ) + (B + C) ( + ) The fractions on the left and right are equal for all values of. The denominators are equal, and so too must be the numerators. So, from just the numerators: A( + ) + (B + C) We can epand the brackets on the right and collect like terms to give: A + A + B + C A + C + (A + B) Now compare the coefficients of the terms on the left- and right-hand sides. Comparing constants: A, and hence A. Comparing coefficients of : 0 C, and hence C 0. Comparing coefficients of : 0 A + B. 9 c mathcentre August 8, 004

10 We already know that A and so B must equal. Returning to the integral, and using these values of A, B and C we find ( + ) d + d The first integral on the right is a standard form. The second is adjusted as follows to make the numerator the derivative of the denominator: + d + d Then ( + ) d + d ln ln + + c Using the laws of logarithms the first two terms can be combined to epress the answer as a single logarithm if required. Eercises 4 ln ln + ln ln + / ln ln + ln +. Find the following integrals by epressing the integrand in partial fractions (a) d (b) ( + 4)( ) ( + )( + + ) d. By completing the square find. Find + + d. + + d. 4. Epress in partial fractions and then integrate Answers ( + )( ). Eercises. (a) ln + ln + +c (b) ln ln 4 +c (c) 5 8 ln +9 8 Eercises. (a) the partial fractions are: 4 ( + ) the integral is ln + ) + ln + C 6 ; ln +7 +c. c mathcentre August 8, 004 0

11 (b) the partial fractions are ( + ) ; the integral is + ln + ln + + C. + (c) the partial fractions are ( 7) 49( 7) ; the integral is 49 ln 8 7( 7) ln 7 + C. 49 Eercises. (a) the partial fractions are ; the integral is ln + C. (b) the partial fractions are ; the integral is + + ln + + C. (c) the partial fractions are 7 + ; the integral is 7 + ln + C. (d) the partial fractions are ; the integral is 7 + ln + + ln + C. Eercises 4. (a) the partial fractions are ; the integral is ln + ln C. (b) the partial fractions are ; the integral is ln ( + + ) ln C (a) tan + C (b) ln + tan ( + ) + C. (c) ln 4 + tan + C. c mathcentre August 8, 004

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