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# All Unit 2 notes MTH 162

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This 29 page Class Notes was uploaded by Jack Magann on Friday February 20, 2015. The Class Notes belongs to MTH 162 at University of Miami taught by Dr. Bibby in Spring2015. Since its upload, it has received 163 views. For similar materials see Calculus 2 in Mathematics (M) at University of Miami.

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Date Created: 02/20/15

Calculus 2 Chapter 58 In previous chapters we were unable to solve such equations as this 62quot 63quot 2 lim x gt0 3 sm2x 2 COSCX 2 Since using the substitution method shows the bottom of the equation to be 0 it s limit would be DNE Does Not Exist However using L Hopital s rule this limit can be solved L Hopital s rule Let f x and gx be differentiable functions If 1 limxa f x limxa gx 0 or 2 limx m f x limx m 906 too Then f x 9 06 limxa imxa if both limits exist gx We can now solve the limits we were not able to in earlier chapters Ex 2x 3x e e 2 o 1 11mx0 smce the 11m1t of both equatlons equals zero we can 3 Sln2x 2 cosx2 0 use L Hopital s rule f x 51 96er 63quot 2 262x 363x g x 2 3 sin2x 2 cosx 2 6 cos2x 2 sinx 2e2x3e3x 23 5 11mx0 Now the subst1tut10n techmque prov1des a real 6 cos2x2 Slnx 60 6 answer Certain problems will require that you make use of L Hopital s rule multiple times EX 1 lim w 3 Now we can use L Hopital s rule x 3x32x 1 oo 6x2 10x oo llmx mo m Slnce 1t stlll equals 00 we use L Hopltal s rule aga1n 12x 10 oo 11mxoo 18x 2 Slnce 1t stlll equals 00 we use L Hop1tal s rule aga1n um 12 2 Xquot 18 18 00 and g are called indeterminate forms however they are not the only ones 0 OIO m Indetermmate Forms 5 0X00 oo oo 00 1 00 What s important to know is that L Hopital s rule only works for 00 and so you have to change the other indeterminate forms into them Ex exe x 2 o 1 11mx0 m 0 Use L Hopltal s rule lim ex e x 0 Use L Ho ital s rulea ain lim exe x 1 9 60 25in2x 0 p g 9 60 4cos2x 2 lntanx oo oo 2 11m x60 1ntan 2x oo 00 SEC2 x lirr1 limx0 cotx sec2 x tan 2x cos2 2x 95 tan 2x Use sin2x 2 sin x cos x so when you cancel everything out you get cos2 x 1 lim 1 9 60 cosx 1 3lim ln2x35x21 oo lim 6x210x 3x47x2 xquot 1n3x47x2 xquot 2x35x21 12x37 Rather then multiplying these expressions out we can just nd the values of the highest degrees of X The limit is then just the fraction that the numbers in front of the highest degrees of X make so 6x210x 3x47x2 18x6 18 lim 11m xquot 2x35x21 12x37 xquot 24x6 24 4 lim 1 1 lim x 0 x gt0 ex1 x x 0 xex x 0 1 ex 0 ex 1 11m gt11m 9 60 xexex 1 0 9 60 xexexex 2 sin 2x 0 5 11m sm 2x csc 3x x60 sin 3x 0 2cos2x 2 lim x 3 cos 3x 3 For the indeterminate forms 1 00 000 let L lim fxgx lnL lnlim Uxgx lim 1anxgx 1 lnL 2 11m gx lnfx 2 11m n11 m 1nfx 1 lim lnL 61 e 906 Ex 1 1 limx0 6quot 3x 10 L 1 lnex3x 4 x so lnL 11mx01nex 3xx 11mx0 9 1393quot 2 E 4 This however is not the answer according to the equation above The full answer is e4 Techniques of Integration Chater 60 1 Division For when there are two polynomials in the fraction and the highest degree of X in the numerator is equal to or greater then that of the bottom Also the equation can not be broken into parts 2 f2x 11 x24 2x2x11 2x2 8 x 3 2x2x11 x 3 f x24 f2 x24 x 3 1 2 3 1 x f2fxz8 fx24 2x lnx 8 tan C 2 Rationalizing Substitution fijr J lett Vx 1 nowputxanddxinterms ofy xt2 1 dx2tdt f f tht t212dtf2t22dt t3 21 vx 13 2 mc 3 Completing the Square f x2x613 dx If the bottom polynomial was in the form of x2 bx or x2 bx the integral could easily be done To do this Beak the last number up for bottom to be perfect square plus or minus removed number x1 x1 dx fx2 6x13 fx 324 Obtain the form x2 bx b2 or x2 bx b2 Now use substitution using the completed square wx 3 dwdx and xw3 f x1 x fW31 x 324 w24 W 4 l 2 1K fwz4fw24 21nw 42tan 2 1nx 6x 13 2tan 1 C Calculus 2 Chapter 62 B Trigonometric Substitution To get the idea of when trig substitution is used look at some past examples d 1f9967 sm 1 C xdx 1 2f fx9 x23dx u9 x2 du 2xdx 3f 2x9 x2dx lfudu 32u C 9 x2C 2 2 2 2 3 f 79sz This one can not be done with any of the methods previously used Problems like number 3 above are when trig substitution is used There are three instances where trig substitution is used 1 Integral involved Vaz u2 Set u asinH so that 9 sin 1 and du acos 9 d9 Place the known values on a right triangle so that Use SOHCAHTOA to relate to right triangle Hypotenuse a opposite side from angle u Pythagoras theorem gives adjacent side Vaz u2 We can then nd that Vaz u2 a cos 9 With these values we can substitute in the original equation Try the one above I xzdx W Letx3sin9 dx3c059 d9 9sin 1 V9 x2 Since cos 9 is adjacent over hypotenuse cos 9 2 Substitute in original equation I dex 3 sin923 cos 9d9 J9 29 d9 9 26d9 W 3 COS 9 s1n cos 99 sin29C 2 Integral involved Vaz u2 set u atan 9 so that 9 tan 1 and du asec2 9 d9 Place the known values on a right triangle Adjacent side a opposite side from angle u Pythagoras theorem gives Hypotenuse Vaz u2 We can nd that Vaz u2 a sec 9 3 Integral involved Vuz a2 set u asecH so that 9 2 sec 1 and du asecH tanH d9 Place the known values on a right triangle Adjacent side a Hypotenuse u Pythagoras theorem gives Opposite side Vuz a2 We can nd that Vuz a2 a tan 9 In all three cases you try to substitute the original equation with these new equations to make the derivative easier to nd Ex dx l f 3 Even though 1t 1s not to the 12 power Tr1g Substitutlon stlll applies 94x2 In this case u 2x 3 tanH 30 x 2 item 9 dx gsec2 9 d9 In this case it may help to draw the triangle and label the sides where Opposite side from angle 2x adjacent side 3 and hypotenuse V9 4x2 3 V94x2 Therefore W cos 9 sec 9 2 V9 4962 2 3sec9 3V9 4962 2 27sec39 Now substitute in these for the original equation f dx Bzfgsec26d6 1f d6 ifcosada 27 94x2 27 sec3 6 2 sec 6 18 1 39 9C 185m 3 f x9dx ux a3 Letx3sec9 dx3sec9tan9 If you draw a triangle Opposite side from angle sz 9 Adjacent side 3 Hypotenuse x Before substitution set the range of the integral to be in terms of 9 so you don t have to change the nal equation to be back in terms of x x3 33sec9 1sec9 90 x6 63sec9 2sec9 9 Substitute everything into the original equation f6Vx2 9dx 13 IStan0 3 x 035ec0 3 sec 9 tan 9d9 f0 3 tan2 9 d9 Now use the Pythagorean identity tan2 9 2 sec2 9 1 f 3tan29d9 3f sec29 1d9 3tan9 6 3 g0 3J n 3f Vx2 4x5 x 1 dx For this multiple methods from previous chapters need to be used Start by completing the square in the bottom x2 4x5 x 1 Vx2 4x5 dx x2 4x41x 221 W Use the W substitution method wx 2 xw2 x 1 fwx 221 dx dx dw Now substitute f W1 dW Vw21 Now split this up into two integrals to solve separately W1 Vw21 f dwf W 1 mdwfmdw The rst can be solved using u substitution f f MI 1 2 2 l d u Wz1dw u w 1 du dew deW Zf u The second is solve using trig substitution 2 W w 1tan9 dwsec29 secH w11Vw21 In triangle Opposite w adjacent 1 and hypotenuse w2 1 2 viz1 fscz9 fseCH d9 1nsec9 tanHI 1nW2 1 WI The nal answer should then be de w21ln w21wC Vx2 4x5 x 1 Vx2 4x5 OI39 dx x 2211n x 221x 2C w21 x 1 f dxxx2 4x5lnx2 4x5x 2C Vx2 4x5 Calculus 2 Chapter 61 Integration By Parts This technique is used to solve such problems as fxcosxdx and flnx The idea is to take one integral and break it up into parts to solve it So recall that f du u And that if v vx and u ux Then it duv udv vdu Know this we can get the integration by parts formula fduv fudv fvdu uvfudvfvdu f udv 2 m7 f vdu formula for integration by parts Ex 1 fxcosx dx Step 1 select a u and a dv and nd du and v ux dudx and dvcosxdx vfcosxdxsinx Step 2 Plug these into the equation above fxcosx dx xsinx fsinx dx xsinx cosx C General Rule dV is usually the most complicated factor of the two that can be integrated using a formula 2 f1nx dx u lnx du dx dv dx 1 x lnxdxxlnx x l dxx1nx xC f f x 3f sin 1x dx u sin 1x du 162 dx dv dx 1 x f sin 1x dx x sin 1x xdx x sin 1x J x1 x2 dx x Now use u substitution but since we already are using a u lets do W substitution w1 x2 dw 2x x sin 1x fx1 x2 dx xsin 1x fw dw 1 3 xsin 1xw C xsin 1x1 x2 C 4 fxzsinx dx ux2 du2xdx dvsinx dx v cosx fx2 sinx dx x2 cosx f Zxcosx dx x2 cosx focosx dx Left with this integration of parts is needs to be done again so u x du dx dv cosxdx v sinx x2 cosx2xsinx fsinxdx x2 cosx2xsinx2cosxC Tabular Method Can only be used when you have fxquot sinax dx fxquot cosax dx or fxneaquot dx For the previous problem of f x2 sin x dx U V39 x2 sin x Step 1 make a quick table 2x cos x 2 sin x Step 2 differentiate u on the left and 0 COS x integrate v on the right until the u gets to zero Step 3 Now pair up the values diagonally so x2 pairs with cosx and so on Step 4 starting at the top line with positive and alternating between that and positive go down the rows as seen above Step 5 Now write out the pairings with their corresponding signs sz sinx dx x2 cosx 2x sinx 2 cosx O C x2 cosx 2x sinx 2 c0506 C More rules Need positive integer power of X The second function needs to be easy to integrate many times Despite these the Tabular method may not be the best depending on the situation Ex 1 f x339 dx Using the tabular method I u v x3 equot x369 39626quot 6xequot 66quot C 3x2 6quot 6x 6quot 6 equot 0 ex 2fx4lnx dx ulnx duidx dvx4 v x5 151 j141511151511 5C 5x nx 5x 5x nx 55x 5x nx 25x 3 fex sinx dx u sinx du cosx dx dv equot v equot equot sinx fex cosx dx u cosx du sinx dv equot v equot equot sinx ex cosx f equot sian dx We got the original function again Rather than integrating again we can use the rule QAB Q so 2QAB Therefore fex sinx dx equot cosx equot sinx fex sinx dx 2 f equot sinx dx equot cosx equot sinx fex Sinx dx ex cosx2ex sinx C 4 fsec3x dx fsecxseczxdx u secx du secxtanxdx dv seczxdx v tanx secxtanx ftanxsecxtanxdx secxtanx ftanzxsecxdx tanzx seczx 1 secxtanx fsec2x 1 secxdx secxtanx f sec3x secxdx 2 sec x tan x f sec3 x dx f sec x dx The original question appears here so 2fsec3xdx secxtanx fsecxdx sec x tan xlnsec xtan x gt fsec3xdx C 2 5 sin In x dx u sinln x du icos n x dv dx 1 x xsinlnx fcoslnx dx u coslnx du sinln x dv dx 1 x x sinln x x cos x f sinln x dx The original question appears here so sinln x x cos x 2 2 f sinln x dx sinln x x cosx gt fsinln x dx C 6 f e dx Use rationalizing substitution from 60 t t2x dx2t fettht2fettdt now the tabular method can be used since it matches f xneaquot dx 2tet et 2 z e 26 C Calculus 2 Chapter 62 A Trig Integrals There are three types Type 1 f sinm x cosn x dx To solve for this one you either use the identity sinzx coszx 1 or on of the power reducing formulas sin2 x 1 cos 2x cos2 x 1 cos 2x To use the power reducing formulas you need to have these special cases a Only f sinm x dx where m is a positive integer b Only f cosn x dx where n is a positive integer c f sinm x cosquot x dx where n and m are positive integers If there is an odd power you need to use the trig identity to solve in terms of either cos x 0quot sin x Ex Special Cases lf sin25x dx since it is to a positive even we can use the power reducing formula 1 1 1 2 5f 1 cos10x dx 5x E sm10x C 2fsin2xcoszx f1 cost ECl cost dx if1 coszx dx 1 1 1 1 1 1 1 1 Zfdx Zf 1 cos4xdx 2x 5x 251n4x 5x 3 Zs1n4x C 3 f cos4 x dx fcos2 xcos2 xdx f1 cos2x 1 cos2x dx if cos2x2 if 2 cos 2x cos2 2xdx separate the integral to simplify if 1 2cos2x dx f 1 cos4xdx 2 ix isin2x x cos4x C Odd Integer Cases 4 f sin4 x cos3 x dx fsin4 x cos2 x cos x dx fsin2 x 1 sin2 x cos x dx f sin4 x sin6 x cos xdx use u substitution and the power rule from here u sinx du cosx dx so fsin4 x sin6 x cos xdx fu u6du u5 u7C sin5x sin7xC 5 fcos10 x sin3 x dx fcos10 x sin2 x sin x dx fcos10 x 1 cos2 x sin xdx fcos10 x cos12 x sin xdx u cosx du sinx 1 1 1 1 fu1 u12 du u11 u13 C cos11x cos13x C 11 13 11 13 Type 2 fsecm x tanquot x dx For this one either you use usecx dusecxtanxdx or utanx dusec2x dx to put the integral in terms of either one of these two trig function Make use of the identity 1 tan2 x 2 sec2 x Type 3 f cscm x cotn x dx For this one either you use u cscx du cscxcotxdx or u cotx du csc2x dx to put the integral in terms of either one of these two trig function Make use of the identity 1 cot2 x csc2 x Ex 1f tan5 x sec x dx Can t use u tan x since there is no sec2 x dx to replace for du so u secx du secxtanxdx gt ftan4 x sec x tan xdx use the identity tan2 x 2 sec2 x 1 gt fsec2 x 12secx tan xdx fsec4 x 2 sec2 x 1secx tan xdx fu4 Zu2 1du 115 u3 u C secsx sec3x secx C 2 f sec6 x dx u tan x this is u because it is easy to break du 2 sec2 x away 2 fsec4 x sec2 xdx Use the identity sec2 x tan2 x 1 ftan2 x 12sec2 xdx ftan4 x 2 tan2 x 1 sec2 xdx fu4 Zu2 1du 115 2113 u C tan5x tan3x tanx C 3 f cot3 x csc3 x dx u can t be cotx since you can t use an identity to put the Whole equation in terms of cotx when du csc2 x is taken out u csc2 x du cotxcscxdx fcot x csc3 x dx f cot2 x csc2 x cotx csc xdx Use the identity cot2 x csc2 x 1 fcsc2 x 1csc2 xcotx csc xdx fcsc4 x csc2 x cotx csc xdx fu4 u2du u5 u3C csc5x csc3xC Calculus 2 Chapter 63 Integration by Partial Fractions To get the idea of partial fractions lets do an algebra 1 problem 3 2 3x 15 2x4 x 19 subtraCt x2 E x2x 5 x2 3x 10 Now lets do a calculus 2 problem using that answer above 96 19 3 2 dx x2 3x 10 f Integrate f 2 x5 Solve 31nx2 21nx 5C This problem was pretty straight forward but what If we don t know f x19 dxf 3 idx x2 3x 10 For the problem above Step 1 Factor the denominator of the integral x 19 x 19 fxz Bx lo fx2x 5 dx Step 2 Set equation in the integral to equal its partial fractions and multiply entire equation by the original denominator 96 19 A B x2x 5 x2 x S gt x 19Ax 5Bx2 If the equation was an improper fraction would would have to divide the numerator by the denominator before setting it equal to its impartial fractions Remember an improper fraction is when the degree of the numerator is equal to or greater then that of the denominator Step 3 Solve for A and B 2 methods Method 1 Substitution Set X equal to a number that eliminates only one of the variables x 2 21A 70 A3 x5 140B7 B 2 Substitute them back into the original equation and solve x 19 i 2 3 i fx2x 5 fx2 x 5 f dx Method 2 Equating the coef cients Multiply out variables and combine like terms x 19Ax 5Bx2 gtx 19xAB 5AZB Set the corresponding coef cients equal to each other ABl 19 5AZB Add the functions together to cancel out one of the variables 5ASB5 B2 A3 SA 23 19 7B 2 14 Step 4 Separate the integral by the partial fractions and solve 95 19 3 i Imam fem x sdquot 3ln39x239 Zlnlx 539 There are four cases or ways of treating the factor in the denominator when splitting the original function into partial fractions 1 NRFL Nonrepeating Linear Factors The function such as x 2 from above has X to the rst degree and the Whole function its self is to the rst degree A x z x2 Only When solving this case use method 1 to solve for the variables 2 RLF Repeating Linear Factors The function has X to the rst degree and the Whole function its self is to the second degree or higher A B A B C x 22 x 2 x 22 quotquot quot or x 23 x 2 m m 3 NRQF Nonrepeating Linear Factors The function has X to the second degree or higher and the Whole function its self is to the first degree AxB x2 2 x2 2 39 39 39 39 quot 4 RQF Repeating Linear Factors The function has X to the second degree or higher and the Whole function its self is to the second degree or higher AxB DxC x2 22 x2 2 92 22 39 39 39 39 quot Here is a practice example for how you would separate a more complex fraction into its partial fractions x92x55x7 A B C D ExF GxH xI x3x 3x24x292 x x2 x3 96 3 x24 x29 x292 EX 3x2x 2 3x2x2 M 1 f x34x x f xx2 4 dx fxx 2x2d 3x2x2 1 i L xx 2x2x x 2 x2 1 3x2x 2Ax 2x2Bxx2Cxx 2 1 x0 2 4A00 A 1 x 2 2008C C Z 3 x2 6O830 B21 fdxI ix2fix iz 11 11 231 2C 2 fix 4 nx 4 nx 2 x x2 dx 2 f x21 Since this is an improper fraction since the degree of x of the numerator is equal to or greater than that of the denominator the functions need to be divided x2x2 x2x2 x3 x2 1 f x21 dx f1dxfx21dx x3 Remainder 1 The second integral now needs partial fractions but the denominator is not a case of NRQF since it can be factored It is rather two cases of NRLF x3 x3 A B x2 1x 1x1 x 1 x1 x3Ax1Bx 1 x1 422A A2 x 1 2 ZB B 1 x21nx 1 lnx1C de dx f1dx f x l x1 3 f3x2 x 1dx f3x2 x 1 x 3x2 x 1 5 3 C x3 x2 x2x 1 x2x 1 x x2 x l 3962 x 1Axquotx 1Bx 1Cx2 3x2 x 12Ax2 AxBx BCx2x2ACx AB B 1 B gt 321 xx AB gt 1 A1A2 3x2x2AC gt 3AC 32c c1 2 x 1 2 1 1 1 j 2 21nx lnx 1C x 96 1 x x2 4x 4 x2 4x 4 A BxC 4 f x 2x24 dx x 2x24 a x24 x2 4x 4Ax24BxCx 2 x2ABx ZBC4A 2C x2x2AB AB1 4xx ZBC 4 ZBC 44A ZB 22A C Solve for A by adding 2nd and 3rd equations and add first 2A ZB 6 2AZB2 gt 4A 4 A 1 Plug in A to solve for the rest of the variables 1 B 1 B 2 4x ZBC C20 x24x4 L 2x 2 fx2x24 dx x2 f2x24 lnlx 2 lnx 4 C x3x2 x3x2 A BxC DxE 5 fxx212 dx 96xz12 x21 x212 x3x2Ax212BxCx3xDxEx gt x3x2Ax42AxABx4Bx2Cx3CDx2Ex gt x4ABCx3x22ABDxCEA AB0 C1 2ABD0 CE1 A2 2B0 4 2D0 1E1 B 2 D 2 E0 jx3x2dx2 Jgdx 2x1dx 2x1 dx Mac2 12 x x2 1 x2 12 Split up the middle integral so it is much easier to solve gt21d I 2x dj1 d I 2x d xx x21x x21x x212x 1 1nle 1 tan 1x m For the last integral remember that f d where u x2 1 Calculus 2 Chapter 63 Integration by Partial Fractions To get the idea of partial fractions lets do an algebra 1 problem 3 2 3x 15 2x4 x 19 subtraCt x2 E x2x 5 x2 3x 10 Now lets do a calculus 2 problem using that answer above 96 19 3 2 dx x2 3x 10 f Integrate f 2 x5 Solve 31nx2 21nx 5C This problem was pretty straight forward but what If we don t know f x19 dxf 3 idx x2 3x 10 For the problem above Step 1 Factor the denominator of the integral x 19 x 19 fxz Bx lo fx2x 5 dx Step 2 Set equation in the integral to equal its partial fractions and multiply entire equation by the original denominator 96 19 A B x2x 5 x2 x S gt x 19Ax 5Bx2 If the equation was an improper fraction would would have to divide the numerator by the denominator before setting it equal to its impartial fractions Remember an improper fraction is when the degree of the numerator is equal to or greater then that of the denominator Step 3 Solve for A and B 2 methods Method 1 Substitution Set X equal to a number that eliminates only one of the variables x 2 21A 70 A3 x5 140B7 B 2 Substitute them back into the original equation and solve x 19 i 2 3 i fx2x 5 fx2 x 5 f dx Method 2 Equating the coef cients Multiply out variables and combine like terms x 19Ax 5Bx2 gtx 19xAB 5AZB Set the corresponding coef cients equal to each other ABl 19 5AZB Add the functions together to cancel out one of the variables 5ASB5 B2 A3 SA 23 19 7B 2 14 Step 4 Separate the integral by the partial fractions and solve 95 19 3 i Imam fem x sdquot 3ln39x239 Zlnlx 539 There are four cases or ways of treating the factor in the denominator when splitting the original function into partial fractions 1 NRFL Nonrepeating Linear Factors The function such as x 2 from above has X to the rst degree and the Whole function its self is to the rst degree A x z x2 Only When solving this case use method 1 to solve for the variables 2 RLF Repeating Linear Factors The function has X to the rst degree and the Whole function its self is to the second degree or higher A B A B C x 22 x 2 x 22 quotquot quot or x 23 x 2 m m 3 NRQF Nonrepeating Linear Factors The function has X to the second degree or higher and the Whole function its self is to the first degree AxB x2 2 x2 2 39 39 39 39 quot 4 RQF Repeating Linear Factors The function has X to the second degree or higher and the Whole function its self is to the second degree or higher AxB DxC x2 22 x2 2 92 22 39 39 39 39 quot Here is a practice example for how you would separate a more complex fraction into its partial fractions x92x55x7 A B C D ExF GxH xI x3x 3x24x292 x x2 x3 96 3 x24 x29 x292 EX 3x2x 2 3x2x2 M 1 f x34x x f xx2 4 dx fxx 2x2d 3x2x2 1 i L xx 2x2x x 2 x2 1 3x2x 2Ax 2x2Bxx2Cxx 2 1 x0 2 4A00 A 1 x 2 2008C C Z 3 x2 6O830 B21 fdxI ix2fix iz 11 11 231 2C 2 fix 4 nx 4 nx 2 x x2 dx 2 f x21 Since this is an improper fraction since the degree of x of the numerator is equal to or greater than that of the denominator the functions need to be divided x2x2 x2x2 x3 x2 1 f x21 dx f1dxfx21dx x3 Remainder 1 The second integral now needs partial fractions but the denominator is not a case of NRQF since it can be factored It is rather two cases of NRLF x3 x3 A B x2 1x 1x1 x 1 x1 x3Ax1Bx 1 x1 422A A2 x 1 2 ZB B 1 x21nx 1 lnx1C de dx f1dx f x l x1 3 f3x2 x 1dx f3x2 x 1 x 3x2 x 1 5 3 C x3 x2 x2x 1 x2x 1 x x2 x l 3962 x 1Axquotx 1Bx 1Cx2 3x2 x 12Ax2 AxBx BCx2x2ACx AB B 1 B gt 321 xx AB gt 1 A1A2 3x2x2AC gt 3AC 32c c1 2 x 1 2 1 1 1 j 2 21nx lnx 1C x 96 1 x x2 4x 4 x2 4x 4 A BxC 4 f x 2x24 dx x 2x24 a x24 x2 4x 4Ax24BxCx 2 x2ABx ZBC4A 2C x2x2AB AB1 4xx ZBC 4 ZBC 44A ZB 22A C Solve for A by adding 2nd and 3rd equations and add first 2A ZB 6 2AZB2 gt 4A 4 A 1 Plug in A to solve for the rest of the variables 1 B 1 B 2 4x ZBC C20 x24x4 L 2x 2 fx2x24 dx x2 f2x24 lnlx 2 lnx 4 C x3x2 x3x2 A BxC DxE 5 fxx212 dx 96xz12 x21 x212 x3x2Ax212BxCx3xDxEx gt x3x2Ax42AxABx4Bx2Cx3CDx2Ex gt x4ABCx3x22ABDxCEA AB0 C1 2ABD0 CE1 A2 2B0 4 2D0 1E1 B 2 D 2 E0 jx3x2dx2 Jgdx 2x1dx 2x1 dx Mac2 12 x x2 1 x2 12 Split up the middle integral so it is much easier to solve gt21d I 2x dj1 d I 2x d xx x21x x21x x212x 1 1nle 1 tan 1x m For the last integral remember that f d where u x2 1

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