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# Math138 - MATH 138 Calculus 2 Fall 2010 Solutions#2 J c I.1 l i = A B I Multipiy both sides(m 1(\$ 2 to get a l = A(\$ 2 5(13 1 Substltutmg xi 33 2:r

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Unformatted text preview: MATH 138 Calculus 2 Fall 2010 Solutions #2 J c I _ .1. \ l} i; = A + B I Multipiy both sides (m +1)(\$ + 2) to get a: -- l = A(\$ + 2} + 5(13 + 1). Substltutmg xi + 33,- + 2 :r + 1 :I‘: + 2 -2 for 2: gives —3 = —B d:- B = 3. Substimting —1'fcn' :1: gives -2 = A. Thus, 1_;”E‘_1._d\$= 1 '2 + 3 )dzz[—2111|m+1|+31nlz+2|]:I D\$E+3I+2 a \$+l 3+2 :[—21n2+31n3)—(—2ln1+31n2)= 31113— 51112 [or in ill”) 333—411—111 3.1—4 3m~4 A B —:: -—-———-—v—‘w' —= ‘ xz—ac—a 3+1+(:r——3)(u:+2) “(x—3H1”) ar—3+m+2 The" 31—4=A(I+2)+B(a¢—3).Takinga:=Iiandr:—2,wcgcl5=5A <¢ A=1and—1=—SB “:- 8:2, so 13 .._ 1 1 [WdIZ/ :I'+l-1'-- 1_+ 2, (1.13: l372‘+5174-111|\$—3l—i—2111(:1:+2) U a“ ~—:.-'—(1 0 1—3; \$+2 2 u =(%+]+1112+21113)——(0+0+ln3+2ln2)=§+1113—1112=%+111% :1. (c1 .1 Jr'—5.:I3+16 A B C => 3:2 — 5:: + 16 = Ame — 2)‘2 + 13(1: — 2x23 +1)+ C(2z + 1). I + (2I+l)(I—2)2—23+l 33—2 (111—2)2 Setting 1: = 2 gives 10 = SC. 50 C = 2. Setting 1: = —% gives = 35114, so A = 3. Equating coefﬁcients off. we get 1=A+2B.SD-—2=QBandB=—1.Tllus, 1r2_5x+15 3 l '2 3 '2 f—{2:B+1){\$_2)2(Ix—/(2x+l —\$_2+(m_2}2)dm—Elnl2m+1|—lnlx—2]—~I_2+C ma+mg+2m+1_ Am+B +CI+D :Lle) \$2 + 2 . Multiply both sides by (:1:2 + 1) (a:2 + 2) to 31:: m # m3 +1 I‘"‘+a“-2 +2ar+1=(Az+B)(z"+2)+(Cz+D)(xz+1) <=> Iii-+32 +2:c+ l = (A33+Bz2+2Ax+QB) + ((31:3 +Dzz+Cz+D) c:- :z:3 + 2:2 + 2x + 1 = (A + 0)::3 + (B + Db”:2 + {2A + (7)3: + (28 + D). Comparing coefﬁcicms gives us the following system of equations: A+C=1 (1] 2A+C=2 (3) B+D=1 (2) 2B+D=1 (4) Subtracting equation (1) from equation (3) gives us A = 1, so C = D. Subtracting equation (2) from equaiion (4} gives us \$3+x2+2r+1 _ a: 1 a: _ 2 B——0.soD—1.Thus.]— —--—-—da:—f +3:2+2 dz.Forjzg+ldx,letu—v:r +1 (\$2 +1)(z2 + 2) m3 +1 3 1 1 1 1 2 f 1 = =— —d =—i C=—l 1 C.F rinse sodu Ezdxandﬂienfm2+1dx 2/” u 2nful+ 2n{:c + )+ or \$2+2 u 1 1 -1 .1: F 110 lb = 2.50f d\$=/———dx=——tan —+C. ormua w: a J- ;n2+2 ISM-(ﬁr ﬁ ﬁ 1 . 1 _1 1: Th ,I=—i :r‘+l +—ta.n ~——+C. us 2 [1(1 ) ﬂ LL‘I a = J? + 3, so u": = 47 + 3 and gudu : (II. The” / “I-L' : ' 2min _ 2” - 2“ 2m:- + .‘i .i- .1: 21L+ (u? _ 3} ‘- /———-—-ug+2u _3 tin =/ (“+3){U_ 1) tin. Now 2|: 1.1 B —_—.___ {u + 3)“: — i) __ r: +3 ” _1 ;:~ ‘2u:,1(u ~1)+ BU: ~.'— 3). Swing u = lngSg : 4350 B : .1. 2. fuming u = —Ii gives —|i 2 —v1.vl. so .-i : Thus. /' Hi: I! ' .' _ (n -+- 15H” *- 1} t H _ (a +3 + u, :. l (in) = 31.. |u +:;r + 5 lulu — 11+ C = %111[\/;r + .1 + :3) + 51:: Mr. + :z — I] + C“ uJ.-' i Let u = 9‘. so that (in = 2” (ix. Then/ 1 :1 But +C (u—2)(u2+1] u-—2+ (3+1 => 1: Ah:2 +1)+{Bu +C)(u — 2). Setting 1; = 2gives 1 = 5A, so A = Setting 1: = 0 gives 1 = g — 26‘. so C = Comparing coefﬁcients of u2 gives 0 = g + B, so 8 = -—%. Thus, l % _-én—§ 1 1 1 u 2 1 [ht—2H1?+l)d‘rm—-/(u—2T u"+l (fa—'3 u—2du—gft131-ldu—5 14:2+1ml Eula-2| — Elnlu") +l| —-%tun"u+C £111 le’ — 2| — % In(e2as + I) — mn'l e" + C _._—-—-—'— ' _ ‘L Let n = 1 + .732, so that (in = 2a: dz. Then at“ 2‘2 f1: -—1 1 1 1/2 —1/2 _1 2 a}: _2_ In C dz: (Ed's): —12—(: du}=— (u -—u )dn— 2 qu u + «1+2? /\/1+:r.'-‘ u" ' 2 ‘ =%(1-+—::12]:W — (1 +33)”: + C [or ﬂat? — 2)x/1+:r2 +0] D ux-H X <5; “‘12. ‘ E (DLM) AJL no 3 d) 3 l : L 1-10 |i (3x. ...
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## This note was uploaded on 11/25/2010 for the course MATH 138 taught by Professor Anoymous during the Spring '07 term at Waterloo.

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Math138 - MATH 138 Calculus 2 Fall 2010 Solutions#2 J c I.1 l i = A B I Multipiy both sides(m 1(\$ 2 to get a l = A(\$ 2 5(13 1 Substltutmg xi 33 2:r

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