Integration that leads to logarithm functions - Mathcentre

Integration that leadsto logarithm functionsmc-TY-inttologs-2009-1Thederivativeof ln xis 1 x . Asaconsequence,ifwereversetheprocess,theintegralof 1 x isln x + c. Inthisunitwegeneralisethisresultandseehowawidevarietyofintegralsresultinlogarithmfunctions.Inordertomasterthetechniquesexplainedhereitisvitalthatyouundertakeplentyofpracticeexercisessothattheybecomesecondnature.Afterreadingthistext,and/orviewingthevideotutorialonthistopic,youshouldbeableto:•recogniseintegralsinwhichthenumeratoristhederivativeofthedenominator.•rewriteintegralsinalternativeformssothatthenumeratorbecomesthederivativeofthedenominator.•recogniseintegralswhichcanleadtologarithmfunctions.1. Introduction 22. Someexamples 3Contentswww.mathcentre.ac.uk 1 c○mathcentre2009

1. IntroductionWealreadyknowthatwhenwedifferentiate y = ln xwefind dydx = 1 x .Wealsoknowthatifwehave y = ln f(x)andwedifferentiateitwefind dydx = f ′ (x)f(x) .Thepointisthatifwerecognisethatthefunctionwearetryingtointegrateisthederivativeofanotherfunction,wecansimplyreversetheprocess.Soifthefunctionwearetryingtointegrateisaquotient,andifthenumeratoristhederivativeofthedenominator,thentheintegralwillinvolvealogarithm:and,reversingtheprocess,if y = ln f(x)sothat∫ f ′ (x)dx = ln(f(x)) + c.f(x)dydx = f ′ (x)f(x)Thisprocedureworksifthefunction f(x)ispositive,becausethenwecantakeitslogarithm.Whathappensifthefunctionisnegative? Inthatcase, −f(x)ispositive,sothatwecantakethelogarithmof −f(x).Then:if y = ln(−f(x))sothatdydx = −f ′ (x)−f(x) = f ′ (x)f(x)and,reversingtheprocess,whenthefunctionisnegative.∫ f ′ (x)dx = ln(−f(x)) + cf(x)Wecancombineboththeseresultsbyusingthemodulusfunction.Thenwecanusetheformulainbothcases,orwhenthefunctiontakesbothpositiveandnegativevalues(orwhenwedon’tknow).Key PointTointegrateaquotientwhenthenumeratoristhederivativeofthedenominator,weuse∫ f ′ (x)dx = ln |f(x)| + c.f(x)www.mathcentre.ac.uk 2 c○mathcentre2009

2. Some examplesExample∫Find tan x dx.Recallthatwecanrewrite tan xas sin x .Observethatthederivativeof cosxis − sin x,sothatcosxthenumeratorisverynearlythederivativeofthedenominator.Wemakeitsobyrewriting sin xcos xas − − sin xcosx andtheintegralbecomes∫∫ sin xtanxdx =cosx dx∫ − sin x= −cosx dx= − ln | cosx| + c.Thisresultcanbewritteninthealternativeform ln | sec x| + c.Example∫xFind1 + x 2dx.Thederivativeofthedenominatoris 2x.Notethatthenumeratorisnotquitethederivativeofxthedenominator,butwecanmakeitsobyrewriting1 1 + x 2as 2 · 2x1 + x 2.ThenExample∫1Findx ln |x| dx.∫x1 + x 2dx = 1 ∫22x1 + x 2dx= 1 2 ln |1 + x2 | + c.Rememberthatthederivativeof ln |x|is 1 x . Sowerewritetheintegrandslightlydifferently:1x ln |x| = 1/xln |x| .Nowthenumeratoristhederivativeofthedenominator.So∫∫1 1/xx ln |x| dx = ln |x| dx= ln |ln |x|| + c.www.mathcentre.ac.uk 3 c○mathcentre2009

Example∫ x cosx + sin xFinddx.x sin xFirstofallthinkaboutwhatwewouldobtainifwedifferentiatedthedenominator:let’sdothisfirst.If y = x sin x,thenusingtheproductruleofdifferentiation,dydx= x cosx + sin x.Soweseethatintheintegralwearetryingtofind, thenumeratoristhederivativeofthedenominator.So∫ x cosx + sin xdx = ln |x sin x| + c.x sin xExercises1.Determineeachofthefollowingintegrals∫∫∫3a)2 + 3x dx b) x1 + 2x dx c) e 2x2 e 2x + 1 dx∫e 2x ∫∫d)e 2x − 1 dx e) x −3cotxdx f)x −2 + 4 dx∫ f′2.Foreachofthefollowingintegrals,usetheresult = ln |f|+ctodeterminetheintegral.fThenrepeattheintegral,usingalgebratosimplifytheintegrandbeforeintegration.Checkthatthetwoanswersobtainedarethesame.∫ 3x2∫ cosxesinx∫ 4x−5a)x dx b) dx c) dx3 e sin x x−4 ∫ x−1/2∫ 5e5x∫ x−5/2d) dx2x e) 1/2e dx f) dx5x x−3/2 Answers1.1a) ln |2 + 3x| + c b)4 ln ∣ ∣ 1 + 2x2 1 + c c)2 ln ∣ e 2x + 1 ∣ + c1d)2 ln ∣ ∣e 2x − 1 ∣ + c e) ln |sin x| + c f) − 1 2 ln ∣ ∣x −2 + 4 ∣ + c2. a) ln ∣ ∣ x3 ∣ ∣ + c = 3 ln |x| + c b) ln∣ ∣e sinx ∣ ∣ + c = sin x + cc) − ln ∣ ∣ ∣x−4 ∣ + c = 4 ln |x| + c d) ln ∣x 1/2 1 + c = ln |x| + c2e) ln ∣ ∣ e5x 2+ c = 5x + c f) −3 ln ∣ ∣ x−3/2 + c = ln |x| + cwww.mathcentre.ac.uk 4 c○mathcentre2009