Integrating algebraic fractions 1


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1 Integrating algebraic fractions mctyalgfrac Sometimes the integral of an algebraic fraction can be found by first epressing the algebraic fractionasthesumofitspartialfractions. Inthisunitwewillillustratethisidea. Wewillsee thatitisalsonecessarytodrawuponawidevarietyofothertechniquessuchascompletingthe 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. Afterreadingthistet,and/orviewingthevideotutorialonthistopic,youshouldbeableto: integrate algebraic fractions by first epressing them in partial fractions integrate algebraic fractions by using a variety of other techniques Contents. Introduction 2 2. Some preliminary results 2. Algebraic fractions with two linear factors 4. Algebraic fractions with a repeated linear factor 6 5. Dealing with improper fractions 7 c mathcentre 2009
2 . Introduction Inthissectionwearegoingtolookathowwecanintegratesomealgebraicfractions.Wewillbe 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 (2 )( + ), 2 + +, ( ) 2 ( + ) and 2 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.itwillnotfactorise),andweconsiderhowtodealwiththiscaseinthesecondvideo on Integrating by Partial Fractions. The third eample contains a factor which is repeated. The fourthisaneampleofanimproperfractionbecausethedegreeofthenumeratorisgreaterthan the degree of the denominator. All of these factors are important in selecting the appropriate way to proceed. Itisalsoimportant toconsider thedegree ofthenumeratorandofthedenominator. For instance, if we consider the third eample, then the degree of its denominator is, because when wemultiplyout ( ) 2 (+)thehighestpowerof is.also,thedegreeofthenumeratoris zero,becausewecanthinkofas 0.Sothedegreeofthenumeratorislessthanthedegree ofthedenominator,andthatisthecaseforthefirstthreeoftheeamples. Wecallfractions liketheseproperfractions.ontheotherhand,inthefinaleample,thedegreeofthenumerator iswhereasthedegreeofthedenominatoris2.thisiscalledanimproperfraction. Key Point Thedegreeofapolynomialepressionin isthehighestpowerof appearingintheepression. Analgebraicfractionwherethedegreeofthenumeratorislessthanthedegreeofthedenominatoriscalledaproperfraction. Ifthedegreeofthenumeratorisgreaterthan,orequalto, the degree of the denominator then the fraction is an improper fraction. 2. 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 2 c mathcentre 2009
3 Alinearfactor, a + binthedenominatorgivesrisetoapartialfractionoftheform A a + b. Repeatedlinearfactors, (a + b) 2 giverisetopartialfractionsoftheform A a + b + B (a + b) 2. Aquadraticfactor a 2 + b + cgivesrisetoapartialfractionoftheform A + B a 2 + b + c. Integration standard results Integration substitution Tofind f () d ln f() + c f() e.g. a 2 + 2d a tan a + c. ( ) 2d,substitute u, du ( ) 2d d ln + + c, + ( ) du dtogive d u 2du u 2 du u + c + c.. Algebraic fractions with two linear factors Inthissectionwewillconsiderhowtointegrateanalgebraicfractionwhichhastheformofa proper fraction with two linear factors in the denominator. Eample Supposewewanttofind (2 )( + ) d. Notethattheintegrandisaproperfraction(becausethedegreeofthenumeratorislessthan the degree of the denominator), and also that the denominator has two, distinct, linear factors. Therefore the appropriate form for its partial fractions is (2 )( + ) A (2 ) + B ( + ) where Aand Bareconstantswhichweshalldetermineshortly. Weaddthetwotermsonthe righthand side together again using a common denominator: (2 )( + ) A (2 ) + B ( + ) A( + ) + B(2 ). (2 )( + ) c mathcentre 2009
4 Becausethefractionontheleftisequaltothatontherightforallvaluesof,andbecausetheir denominators are equal, then their numerators too must be equal. So, from just the numerators, A( + ) + B(2 ). () Wenowproceedtofindthevaluesoftheconstants Aand B. Wecandothisinoneoftwo ways,orbymiingthetwoways. Thefirstwayistosubstituteparticularvaluesfor. The second way is to separately equate coefficients of constant terms, linear terms, quadratic terms etc.bothofthesewayswillbeillustratednow. Substitution of particular values for Becauseepression()istrueforallvaluesof wecansubstituteanyvaluewechoosefor. Inparticular,ifwelet 2thesecondtermontherightbecomeszero,andeverythinglooks simpler: 2 A(2 + ) + 0 fromwhich 5A 2andso A 2 5. Similarly, substituting in epression() makes the first term zero: 5B from which B 5. Thus the partial fractions are (2 )( + ) 2 5(2 ) 5( + ). Bothofthetermsontherightcanbeintegrated: ( ) 2 5(2 ) d 2 5( + ) 5 2 d 5 + d 2 5 ln 2 ln + + c. 5 Notethatinthefirstofthetwointegrals,wehavesetthenumeratortobe andcompensated forthisbywritingaminussignoutsidetheintegral. Wehavedonethisbecausethederivative of 2 is,sothattheintegralisinastandardform.sobyusingpartialfractionswehave broken down the original integral into two separate integrals which we can then evaluate. Equating coefficients Asecondtechniqueforfinding Aand Bistoequatethecoefficientsofequivalenttermsoneach side.firstofallweepandthebracketsinequation()andcollecttogetherliketerms: Equating the coefficients of on each side: A + A + 2B B (A B) + A + 2B. A B. (2) 4 c mathcentre 2009
5 Equating constant terms on each side of this epression gives 0 A + 2B. () Thesearetwosimultaneousequationswecansolvetofind Aand B.MultiplyingEquation(2) by2gives 2 2A 2B. (4) Now,adding()and(4)eliminatesthe B stogive 2 5A fromwhich A 2 5. Also,from(2), B A justasweobtainedusingthe methodofsubstitutingspecificvaluesfor. Oftenyouwillfindthatacombinationofboth techniques is efficient. Eample 2 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 Aand Bareconstantsweneedtofind. Weaddthetwotermsontherighthandside together again using a common denominator. A ( + ) + B ( + ) A( + ) + B. ( + ) Becausethefractionontheleftisequaltothatontherightforallvaluesof,andbecausetheir denominators are equal, then their numerators too must be equal. So, from just the numerators, A( + ) + B. Ifwesubstitute 0wecanimmediatelyfind A: sothat A. Ifwesubstitute wefind B: sothat B.Then 2 ( + ) d A(0 + ) + B(0) A( + ) + B( ) 2 ( ) d + [ ln ln + ] 2 ( ln2 ln) ( ln ln2) 6 ln2 ln ln 26 ln c mathcentre 2009
6 Eercises. Find each of the following integrals by epressing the integrand in partial fractions. + 2 (a) d (b) d (c) ( + 2)( + ) (2 + )( 4) ( )( + 7) d 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 ( ) 2 ( + ) d. InthisEamplethereisarepeatedfactorinthedenominator. Thisisbecausethefactor appearstwice,asin ( ) 2.Wewrite ( ) 2 ( + ) A + B ( ) + C 2 + A( )( + ) + B( + ) + C( )2 ( ) 2 ( + ) Asbefore,thefractionsontheleftandtherightareequalforallvaluesof.Theirdenominators areequalandsowecanequatethenumerators: A( )( + ) + B( + ) + C( ) 2. () Substituting inequation()gives 2B,fromwhich B 2. Substituting gives 4Cfromwhich C 4. Knowing Band C,substitutionofanyothervaluefor willgivethevalueof A.Foreample,if welet 0wefind A + B + C andso A fromwhich A.Alternatively,wecouldhaveepandedtherighthandsideofEquation(), 4 collected like terms together and equated coefficients. This would have yielded the same values for A, Band C. The integral becomes ( ) ( ) 2 ( + ) d 4( ) + 2( ) + d 2 4( + ) 4 ln 2( ) + ln + + c. 4 Using the laws of logarithms this can be written in the following alternative form if required: 4 ln + 2( ) + c c mathcentre 2009
7 Eercises 2. Integrate each of the following by epressing the integrand in partial fractions (a) d (b) d (c) ( + ) 2 ( ) ( + 2) 2 ( + ) ( 7) 2d. 5. Dealing with improper fractions Whenthedegreeofthenumeratorisgreaterthanorequaltothedegreeofthedenominator thefractionissaidtobeimproper. Insuchcasesitisfirstnecessarytocarryoutlongdivision as illustrated in the net Eample. Eample Find 2 4 d. Thedegreeofthenumeratorisgreaterthanthedegreeofthenumerator. Thisfractionis therefore improper. We can divide the denominator into the numerator using long division of fractions: ) so that Notethatthedenominatorofthesecondtermontherighthandsideisthedifferenceoftwo squaresandcanbefactorisedas 2 4 ( 2)( + 2).So, ( 2)( + 2) A 2 + B + 2 A( + 2) + B( 2). ( 2)( + 2) Asbefore,thefractionsontheleftandontherightareequalforallvaluesof.Theirdenominatorsarethesame,andsotoomustbetheirnumerators. Soweequatethenumeratorsto give 4 A( + 2) + B( 2). Choosing 2wefind 8 4Asothat A 2. Choosing 2gives 8 4Bsothat B 2.Sowiththesevaluesof Aand Btheintegralbecomes ( 2 4 d ) d + 2 Eercises ln ln c. 7 c mathcentre 2009
8 . Use long division and partial fractions to find the following integrals (a) 2d (b) d (c) d (d) d Answers Eercises. (a) ln + ln c (b) 22 ln c. 9 8 Eercises 2.(a)thepartialfractionsare: 4 + theintegralis 4 (b) the partial fractions are ( + ) ln + ) + ln + C 6 ( + 2) ; theintegralis + ln + 2 ln + + C. + 2 (c) the partial fractions are ( 7) 2 49( 7) ; theintegralis 49 ln 8 7( 7) ln 7 + C. 49 Eercises.(a)thepartialfractionsare ; theintegralis 2 ln + C. 2 (b)thepartialfractionsare ; theintegralis ln + + C. 2 (c)thepartialfractionsare 7 + ; theintegralis 7 + ln + C. (d)thepartialfractionsare ; theintegralis 7 + ln + + ln + C. 4 ln ln 4 + c (c) 5 ln ; 8 c mathcentre 2009
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