ࡱ>  QuA ebjbj .hhE7  II_w8t7Dk4r{c"y"y""j#.## 63838383838383$58\3IB#6j#j##$$\3II""%4+++&$`I"I"63+#63+++"a(+"3;40k4+9(9+9H+##+#####\3\3*###k4####9######### : Integration It might appear to be a bit obvious but you must remember all of your C3 work on differentiation if you are to succeed with this unit. Content By the end of this unit you should be able to: Integrate trigonometric, exponential functions and  EMBED Equation.3  Find volumes of revolution Find integrals using partial fractions and integrating by parts. Finding solutions to seperable first order differential equations. Find areas under curves and volumes generated from functions given in parametric form. Use of trapezium rule to provide approximate solution to integrals. Standard Integrals Function (f(x))Integral EMBED Equation.3 xn EMBED Equation.3  EMBED Equation.3  EMBED Equation.3 (ax + b)n EMBED Equation.3  EMBED Equation.3  EMBED Equation.3 exex + cSin x-Cos x + cCos xSin x + cTan xln|(Sec x)| + cCot xln|(Sin x)| + cCosec x-ln|(Cosec x + Cot x)| + cSec xln|(Sec x + Tan x)| + cSec2xTanx + cSec x Tan x Sec x + c-Cosec2xCot x + ceax+b EMBED Equation.3 Cos (ax + b) EMBED Equation.3 SinnxCosx EMBED Equation.3 CosnxSinx EMBED Equation.3 Sin2x EMBED Equation.3 Cos2x EMBED Equation.3 Integrals in red are given in formula book (but you should be able to prove them). Integration of trigonometric functions. To succeed with these types of questions you need to have a very good grasp of the trig identities from C3 and obviously the differentiation from the same unit. Integration questions involving trigonometry are either of a substitution type or areas and volumes from parametrics. The following examples give a flavour but cant cover every eventuality. Example 1 By using the substitution u = sinx or otherwise, find  EMBED Equation.DSMT4  Giving your answer in terms of x. From C3 you should remember that: sin2x = 2sinx cosx Hence:  EMBED Equation.DSMT4  If u = sinx du = cosdx (part of the integral!) Therefore  EMBED Equation.DSMT4   EMBED Equation.DSMT4  Example 2 Use the substitution x = sin  to find the exact value of  EMBED Equation.DSMT4  If x = sin  then the denominator becomes (1  sin2) which is cos2. When this is raised to the power of 1.5 the result is cos3 as a denominator. The x s have now been substituted but you can t forget about the dx. If x = sin  then dx = cosd. Therefore the integral becomes:  EMBED Equation.DSMT4  The next thing to spot is that  EMBED Equation.DSMT4 . Therefore the question becomes:  EMBED Equation.DSMT4  The question isn t quite finished as the original limits were given in terms of x. So for the lower limit of x = 0, sin  = 0, hence  = 0. For the upper limit x = 0.5, so sin  = 0.5, hence  = EMBED Equation.DSMT4 . The question asks for the exact value so the integral is finally.  EMBED Equation.DSMT4  There are a few tricks in this question but practice makes perfect and you must remember all of your differentiation and trig work from C3 and C4. Use ideas outlined in the examples above to carry out the next question. Use the substitution u2 = 1 cos2 and integration to find  EMBED Equation.DSMT4 . (You will need to use implicit differentiation when dealing with the u2.) Volume of Revolution The basic principle here is that we take very thin strips, of width x (delta x), underneath a curve in the xy plane and rotate them around the x axis to generate a volume.  The diagram on the left shows the graph of y = x2 in 3D. The pink section is the area under the graph between the x values of 0 and 1. This pink section has been rotated through 360 degrees about the x axis to form the green solid in the diagram on the right. If you look at the far end of the green solid it has a circular appearance. The area of the end is y2 and therefore the volume of a very thin disc of width x will be y2x. We use integration to add up the volumes of all of the very thin discs to give the volume of revolution. " y2x =  EMBED Equation.DSMT4  Curves can be rotated about the x or y axis and therefore there are two formulae to calculate volumes. The volume generated by a curve as it rotates around the x axis is given by:  EMBED Equation.DSMT4  The volume generated by a curve as it rotates around the y axis is given by:  EMBED Equation.3  The example outlined below is very standard but hopefully the pitfalls are well signposted. Example 3 The finite region bounded by the curve with the equation y = 8x x2 and the x axis is rotated through 360 about the x axis. Using integration find, in terms of , the volume of its solid form. The volume generated by a curve as it rotates around the x axis is given by:  EMBED Equation.DSMT4  The mistake a number of candidates make is to state that y2 in this case is 64x2 + x4 WRONG Y2 = (8x x2)( 8x x2) = 64x2 16x3 + x4 Therefore volume is given by:  EMBED Equation.DSMT4  The quadratic, y = 8x x2, has solutions of x = 0 and x = 8 and therefore these become the limits on the integral.  EMBED Equation.DSMT4  Example 4 Find the volume generated when the region bounded by the curve with equation  EMBED Equation.DSMT4 , the x axis and the lines x =  EMBED Equation.DSMT4  and x = 3 is rotated through 360 degrees about the x axis. Volume =  EMBED Equation.DSMT4   EMBED Equation.DSMT4  Take care with the negative powers when integrating and the rules of logs. In most cases the question will specifically ask for answers in terms of a + bln9 for example. A calculator answer will loose you the final answer mark. Integrating by Parts. This concept is guaranteed to appear on a C4 paper and is not too tricky. Integration by parts is used to integrate functions that are made up of products of functions. Not surprisingly the process is the reverse of differentiating a product. The idea is best explained through an example. Example 5 Use integration by parts to find the exact value of  EMBED Equation.DSMT4  The rule for integrating by parts is:  EMBED Equation.DSMT4  The thing to remember is that one part of the question becomes u and the other part becomes EMBED Equation.DSMT4 . Since we cant integrate lnx, it must be u and therefore x5 must equal EMBED Equation.DSMT4 . On the right of the rule we need v and  EMBED Equation.DSMT4  this can be achieved by integrating and differentiating respectively. Therefore:  EMBED Equation.DSMT4  Let u = lnx and  EMBED Equation.DSMT4   EMBED Equation.DSMT4   EMBED Equation.DSMT4  So by substituting into the rule:  EMBED Equation.DSMT4   EMBED Equation.DSMT4  I was telling a little lie before. lnx can be integrated and you should know how!! Differential Equations Most of the differential equation questions will require a number of integration techniques. Integration of trig functions, use of partial fractions or integration by parts could be used. The following example uses integration by parts to find the general solution. Example 6 Given that y = 7 at x = , solve the differential equation  EMBED Equation.DSMT4  Differential equations in C4 are solved by seperating variables (x s and dx to one side, y s and dy to the other). So the question becomes:  EMBED Equation.DSMT4  The left hand side is simply a natural log answer. The right hand side, however, will require two uses of integration by parts and a question of this difficultly has been on recent exam papers. Here goes! If we start by integrating the right hand side by parts:  EMBED Equation.3  We have chosen the x2 to be u because it is made simpler by differentiation.  EMBED Equation.3  Substituting the values into the equation for integrating by parts gives:  EMBED Equation.3  Unfortunately the integral on the right is still too complex so we have to integrate by parts again. At least this time the 2x will differentiate to 2.  EMBED Equation.3  Great care must be taken with the signs in a question of this complexity. The following example tests the concept of partial fractions but then uses the solution in a differential equation. Remember that separation of the variables is the key. Example 7 (a) Express  EMBED Equation.3  in partial fractions. (b) Given that y = 4 at x = 2, use your answer to part (a) to find the solution of the differential equation  EMBED Equation.3  =  EMBED Equation.3 , x > 1.5 Express your answer in the form y = f(x). (a) Express  EMBED Equation.3  in partial fractions.  EMBED Equation.3  By letting x = -1 and x = 0 we find that A = 4 and B = -3. Therefore:  EMBED Equation.3  (b) Given that y = 4 at x = 2, use your answer to part (a) to find the solution of the differential equation  EMBED Equation.3  =  EMBED Equation.3 , x > 1.5 Express your answer in the form y = f(x). As mentioned earlier we solve first order differential equations by separating the variables and then integrate with respect to x and y.  EMBED Equation.3   EMBED Equation.3  If we use the initial condition of y = 4 and x = 2 to find c.  EMBED Equation.3  Therefore:  EMBED Equation.3  We can now remove the logarithms to give the answer in the form y = f(x):  EMBED Equation.3  The following example deals with exponential decay, a regular differential equation question. Example 8 A radioactive substance has a half life of 1200 years (the time taken for half of the substance to decay) and at time t = 0, 6kg are present. If the substance decays at a rate proportional to the amount, x, present at a time t find an equation for x in terms of t. Find also the amount of substance left after 800yrs. So the rate of change is proportional to the amount of the substance hence:  EMBED Equation.DSMT4  Therefore:  EMBED Equation.DSMT4  The negative is used to highlight decay. By separating the variables we get:  EMBED Equation.DSMT4  And integrating gives: lnx + lnC = -kt lnCx = -kt Removing the logs: Cx = e-kt let  EMBED Equation.DSMT4  So finally x = Ae-kt Using the initial conditions of t = 0 and x = 6 A = 6 By definition after 1200 yrs the mass of substance should be 3kg. Therefore: x = 6e-1200k 3 = 6e-1200k Taking ln of both sides: ln0.5 = -1200k k = 0.000578 Hence x = 6e-0.000578t If t = 800 the amount of substance left is: x = 6e-0.000578 800 x = 3.78kg More complex integrations by substitution. Example 9 Use the substitution u2 = 2x 1, or otherwise, find the exact value of  EMBED Equation.DSMT4  From an earlier example you should remember that there are three parts that need dealing with, 6x, the dx and the square root in the denominator If u2 = 2x 1 then 2udu = 2dx ( by implicit differentiation). So dx = udu. If u2 = 2x 1 then x = 0.5u2 + 0.5. The integral becomes:  EMBED Equation.DSMT4  And now for the limits. If x = 13 then u = 5 and if x = 25 then u = 7.  EMBED Equation.DSMT4  Integration of Parametrics to Find Areas under curves. Example 10  The diagram shows a sketch of the curve C with parametric equations x = 7tsint, y = 5sect, 0 d" t <  EMBED Equation.DSMT4  The point P(a, 10) lies on C. a) Find the exact value of a. The region R is enclosed by the curve C, the axes and the line x = a. b) Show that the area of R is given by  EMBED Equation.DSMT4  c) Find the exact value of the area R. a) Find the exact value of a. The point P has y coordinate 10. Therefore: 5sect = 10 If t =  EMBED Equation.DSMT4  sect = 2 x =  EMBED Equation.DSMT4  cost = 0.5 x =  EMBED Equation.DSMT4  t =  EMBED Equation.DSMT4  a =  EMBED Equation.DSMT4  b) When integrating to find the area under a parametric curve you need to integrate the following:  EMBED Equation.DSMT4  The x part of the parametric is a product and therefore we need t   R     Q R f ~  ɾɲ~mbVJ;jh0h; OJQJUh0h; 5OJQJh h; 5OJQJh h])OJQJ!jh hA EHOJQJUj|~8vx`d ^`gdUSbgdUSb2468:<JNPjl2DFtvxzǻvjjj\K!jIh hUSbEHOJQJUjO'I h hUSbUVh hUSbH*OJQJ!jEh hUSbEHOJQJUjW'I h hUSbUVjh hUSbOJQJUh hUSbOJQJh hl5OJQJh hUSb5OJQJh h!OJQJ h hUSbjh hUSbUjBh hUSbEHUjFjI h hUSbUV>@nprt8Cgh`帧噈|n]Rh hlOJQJ!j{Xh hUSbEHOJQJUj)I h hUSbUVh hUSb5OJQJ!jUh hUSbEHOJQJUj;)I h hUSbUV!jRh hUSbEHOJQJUj>(I h hUSbUV!jNh hUSbEHOJQJUj(I h hUSbUVh hUSbOJQJjh hUSbOJQJU`a4n.DGMlnEygd]} ^`gd]}gda[gdUSb`awx246dfhjln,.-BCDEFGwϨ}tl]Nljh bh"OJQJUjah bh"OJQJUhL}OJQJh# 5OJQJh h# 5OJQJh[h[OJQJh[h[H*OJQJh# OJQJh[OJQJ!j]h h# EHOJQJUjt̎J h h# UVjh h# OJQJUh hlH*OJQJh hlOJQJh h# OJQJwxtprvxDE㽲벽vgV!jlh]}h]}EHOJQJUj cK h]}CJUVaJjh]}OJQJU!jih h]}EHOJQJU!j^fh h]}EHOJQJUjI h h]}UVh h]}OJQJjh h]}OJQJUh]}h]}OJQJhL}h]}H*OJQJh]}OJQJhL}OJQJhL}hL}H*OJQJ!yz (!*!2!4!b!d!f!h!!!"" 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h|X)OJQJjh h|X)OJQJU!jh hEHOJQJUAA?@_` )*45gda[ ^`gdo differentiate the first (7t) and multiply it second (sint) then add the first multiplied by the differential of the second. This gives:  EMBED Equation.DSMT4  So to find the area R:  EMBED Equation.DSMT4  The limits are defined from the work in part (a). c) The integral from part (b) becomes:  EMBED Equation.DSMT4  The integral of tant should be in your notes. The denominator nearly differentiates to give the numerator. Therefore the integral is given by the negative ln of cost.  EMBED Equation.DSMT4  = 43.5 If you are asked to find the volume generated by a parametric as it rotates about the x axis then the integral to evaluate is:  EMBED Equation.DSMT4  Example 11 The diagram below is the graph of the function y = Sec x  Use the trapezium rule with five equally spaced ordinates to estimate the area of the region bounded by the curve with equation y = sec x, the x-axis and the lines x = - and x= , giving your answer to two decimal places. The five x coordinates are  EMBED Equation.3 . The best way to succeed with these questions is to put the information into a table. x EMBED Equation.3  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3 Y21.15511.1552Trapezium rule is introduced in AS.  EMBED Equation.3   EMBED Equation.3  Finally an integration question that has an exponential substitution! Example 12 Use the substitution u = 3x to find the exact value of  EMBED Equation.3  The integral very quickly becomes  EMBED Equation.3  but we havent dealt with the dx. From example 9 of the C4 differentiation notes the differential of u is:  EMBED Equation.3  Therefore  EMBED Equation.3  and the integral becomes:  EMBED Equation.3  Remember that the question said exact answer so leave it in terms of logs.     C4  PAGE 1 This principle applies to all trig functions. You should be able to do this as a substitution if you cant learn it by rote. We integrate to find v. 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T5J ս}l0O/!#)%QrQep+"W~"((Us:}R޼6|OZ>44(]]ųgݡG,Qas͇G@tϦM{IbNkj !r6K` u(yiaf3oh(Dd lb # c $A!? ?3"`?"21ה76wu0‡60 >~r `!ה76wu0‡60RxڥRO@.g$'HLvR@L,6KPrZOd*7010SU ;;TU>{;q@D Ƞ98c)b|0(ͤI>+cEBG4b eՉ4*X<y3{in(T$TbKm6~ޯ$%,0ʭmUS8?~뮍?Jt='dL] /Ole -CompObj.iObjInfo0Equation Native 1I¦<DSMT5WinAllBasicCodePagesComic Sans MSSymbolTimes New RomanCourier NewMT Extra!/ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A  p12++ 5x() 2 dx  133 +" ==p14++ 20x++ 25x 2 ()dx  133 +" ==4x++20lnx"- 25x[]  133 ==12++20ln3"- 253"- 43"-20ln 13++75== 2323++20ln9 FMathType 5.0 Equation MathType EFEquation.DSMT49qA%<DSMT5WinAllBasicCodePagesComic Sans MSSymbolTimes New RomanCourier NewMT Extra!/ED/APG_APAPAE%B_AC_AE*_HA@_1233589434FTTOle ACompObjBiObjInfoDEquation Native EA_1233591605FTTOle KCompObjLiAHA*_D_E_E_A  1x 5 lnxdx 13 +" FMathType 5.0 Equation MathType EFEquation.DSMT49qAs<DSMT5WinAllBasicCodePagesObjInfoNEquation Native O_1233591733FTTOle VComic Sans MSSymbolTimes New RomanCourier NewMT Extra!/ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A  u dvdxdx==uv"-v dudx  +" dx  +" FMathType 5.0 Equation MathTyCompObjWiObjInfoYEquation Native Z_1250860461,;FTTpe EFEquation.DSMT49qA<DSMT5WinAllBasicCodePagesComic Sans MSSymbolTimes New RomanCourier NewMT Extra!/ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A   dvdxOle _CompObj`iObjInfobEquation Native c FMathType 5.0 Equation MathType EFEquation.DSMT49qW<DSMT5WinAllBasicCodePagesComic Sans MSSymbolTimes New RomanCourier NewMT Extra!/ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A   dudx_1250860587FTTOle hCompObjiiObjInfok FMathType 5.0 Equation MathType EFEquation.DSMT49qW,<DSMT5WinAllBasicCodePagesComic Sans MSSymbolTimes New RomanCourier NewMT Extra!/ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A  1x Equation Native lH_1233591858FTTOle rCompObjsi5 lnxdx 13 +" == FMathType 5.0 Equation MathType EFEquation.DSMT49qA<DSMT5WinAllBasicCodePagesComic Sans MSSymbolTimes New RomanCourier NewMT ExtrObjInfouEquation Native v,_1233591883FTTOle {a!/ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A   dvdx==x 5 FMathType 5.0 Equation MathType EFEquation.DSMT49qA<DSMT5WinAllBasicCodePagesCompObj|iObjInfo~Equation Native ,_1233591914FTTComic Sans MSSymbolTimes New RomanCourier NewMT Extra!/ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A   dudx== 1x FMathType 5.0 Equation MathType EFEquation.DSMT49qOle CompObjiObjInfoEquation Native #A<DSMT5WinAllBasicCodePagesComic Sans MSSymbolTimes New RomanCourier NewMT Extra!/ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A  v== x 6 6 FMathType 5.0 Equation MathType EFEquation.DSMT49q_1233994915g!FTTOle CompObjiObjInfo}<DSMT5WinAllBasicCodePagesComic Sans MSSymbolTimes New RomanCourier NewMT Extra!/ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A  1x 5 lnxdx 13 +" == x 6 6"-1 1x x 6 6Equation Native _1233590874SFTTOle CompObji 13 +" dx== x 6 6"-1 x6 5 dx== 13 +"  x 6 6"- x 6 36[] 13 == 5x 6 36[] 13 == 4054"- 536== 9109 FMathType 5.0 Equation MathType EFEquation.DSMT49qAa<DSMT5WinAllBasicCodePagesComic Sans MSSymbolTimes New RomanCourier NewMT Extra!/ED/APG_APAPAE%B_AC_AE*_HA@ObjInfoEquation Native }_1233592842FTTOle AHA*_D_E_E_A   dydx==2yx 2 cosx,    y > 0 FMathType 5.0 Equation MathType EFEquation.DSMT49qA[<DSMT5WinAllBasicCodePagesCompObjiObjInfoEquation Native w_1264782652FTTComic Sans MSSymbolTimes New RomanCourier NewMT Extra!/ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A   dy2y  +" ==x 2 cosx  +" dx FMicrosoft Equation 3.0 DS Equation Equation.39qOle CompObjfObjInfoEquation Native d 12lny=x 2 cosxdx  +"       u = x 2    dvdx = cosx  FMicrosoft Equation 3.0 DS Equation Equation.39q_1264783380FTTOle CompObjfObjInfo D 12lny=x 2 cosxdx  +"       u = x 2    dvdx = cosx                                    dudx=2x   Equation Native _1264783140FTTOle CompObjf v = sinx FMicrosoft Equation 3.0 DS Equation Equation.39qud 12lny=uv"vdudxdx  +" 12lny=x 2 ObjInfoEquation Native _1264783884FTTOle sinx"2xsinxdx  +" FMicrosoft Equation 3.0 DS Equation Equation.39qnX6d 12lny=x 2 sinx"2xsCompObjfObjInfoEquation Native _1264850363FTTinxdx  +" 12lny=x 2 sinx"....      u = 2x      dvdx=sinx                                  dudx=2       v = -cosx12lny=x 2 sinx""2xcosx""2cosxdx  +" []12lny=x 2 sinx+2xcosx"2sinx+c FMicrosoft Equation 3.0 DS Equation Equation.39qOle CompObjfObjInfoEquation Native D†  13"2x(2x"3)(x+1) FMicrosoft Equation 3.0 DS Equation Equation.39q_1264850373FTTOle CompObjfObjInfoEquation Native >_1264850369FTTOle CompObjfD"`Y dydx FMicrosoft Equation 3.0 DS Equation Equation.39qD–X y(13"2x)(2x"3)(x+1)ObjInfoEquation Native _1264850379 FTTOle  FMicrosoft Equation 3.0 DS Equation Equation.39qD†4 13"2x(2x"3)(x+1)CompObjfObjInfoEquation Native _1233994990FTTOle CompObjiObjInfoEquation Native W     #$'*+,-./0123456789:;<=>?BEFGHKNOPQRSTUVWZ]^_`abcdefghilopqrstwz{|} FMathType 5.0 Equation MathType EFEquation.DSMT49q};<DSMT5WinAllBasicCodePagesComic Sans MSSymbolTimes New RomanCourier NewMT Extra!/ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A   13"-2x(2x"-3)(x++1)a" A2x"-3++ Bx++113"-2x==A(x++1)++B(2x"-3) FMathType 5.0 Equation MathType EFEquation.DSMT49q_1233995639 FTTOle  CompObj  iObjInfo }DSMT5WinAllBasicCodePagesComic Sans MSSymbolTimes New RomanCourier NewMT Extra!/ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A   13"-2x(2x"-3)(x++1)a" 42x"-3"-Equation Native _1264850395FTTOle CompObjf 3x++1 FMicrosoft Equation 3.0 DS Equation Equation.39qD"pĤ dydx FMicrosoft Equation 3.0 DS EqObjInfoEquation Native >_1264850391FTTOle CompObjfObjInfo!Equation Native "_1233995651FTTuation Equation.39qD–XĤ y(13"2x)(2x"3)(x+1) FMathType 5.0 Equation MathType EFEquation.DSMT49qOle %CompObj&iObjInfo(Equation Native )}r<DSMT5WinAllBasicCodePagesComic Sans MSSymbolTimes New RomanCourier NewMT Extra!/ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A   dydx== y(13"-2x)(2x"-3)(x++1)Seperate variables and integrate: dyy== (13"-2x)dx(2x"-3)(x++1)  +"  +" Using part (a) and integrating the left we get:lny= 42x-3dx"- 3x++1  +"  +" lny==2ln(2x"-3)"-3ln(x++1)++c_1264784035FTTOle @CompObjAfObjInfo C FMicrosoft Equation 3.0 DS Equation Equation.39qd lny=42x"3  +" dx"3x+1dx  +" lny=2ln(2x"3)"3ln(x+1)+Equation Native D_12339959475#FTTOle ICompObj"$Jic FMathType 5.0 Equation MathType EFEquation.DSMT49q}{<DSMT5WinAllBasicCodePagesComic Sans MSSymbolTimes New RomanCourier NewMT Extra!/ED/APG_APAPAE%B_AC_AE*_HA@ObjInfo%LEquation Native M_1233996200(FTTOle XAHA*_D_E_E_A  ln4==2ln(1)"-3ln(3)++cRemember that ln1==0 and 3ln3=ln27Therefore:c = ln4 +ln27=ln108 FMathType 5.0 Equation MathType EFEquation.DSMT49q}'<DSMT5WinAllBasicCodePagesComic Sans MSSymbolTimes New RomanCourier NewMT Extra!/ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A  lnyCompObj')YiObjInfo*[Equation Native \C_1233996208&0-FTT==2ln(2x"-3)"-3ln(x++1)++ln108Using rules of logslny==ln(2x"-3) 2 "-ln(x++1) 3 ++ln108lny==ln 108(2x"-3) 2 (x++1) 3 () FMathType 5.0 Equation MathType EFEquation.DSMT49q}j<DSMT5WinAllBasicCodePagesComic Sans MSSymbolTimes New RomanCourier NewMT ExtrOle jCompObj,.kiObjInfo/mEquation Native na!/ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A  y== 108(2x"-3) 2 (x++1) 3 () FMathType 5.0 Equation MathType EFEquation.DSMT49q_12340043262FTTOle uCompObj13viObjInfo4x<DSMT5WinAllBasicCodePagesComic Sans MSSymbolTimes New RomanCourier NewMT Extra!/ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A   dxdt"x FMathType 5.0 Equation MathTyEquation Native y_1234004372+ 7FTTOle ~CompObj68ipe EFEquation.DSMT49q <DSMT5WinAllBasicCodePagesComic Sans MSSymbolTimes New RomanCourier NewMT Extra!/ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A   dxdt=="-kxObjInfo9Equation Native &_1234004476<FTTOle CompObj;=iObjInfo>Equation Native P_1234004705:AFTT FMathType 5.0 Equation MathType EFEquation.DSMT49q4(<DSMT5WinAllBasicCodePagesComic Sans MSSymbolTimes New RomanCourier NewMT Extra!/ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A   dxx=="-kdt  +"  +" FMathType 5.0 Equation MathType EFEquation.DSMT49q<DSMT5WinAllBasicCodePagesComic Sans MSSymbolTimes New RomanCourier NewMT Extra!/ED/APG_APAPAE%B_AC_AE*_HA@Ole CompObj@BiObjInfoCEquation 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