数学分析讲义第四版刘玉琏傅沛仁著高等教育出版社课后答案第七单元.pdf

8 111?1 ?SSS1.1(1)?SSS1.2(5) ?SSS1.3(9) 111?444? 13 ?SSS2.1(13)?SSS2.2(21) ?SSS2.3(36)?SSS2.4(40) ?SSS2.5(51) 111nnn?YYY 57 ?SSS3.1(57)?SSS3.2(64) 111ooo?YYY555 75 ?SSS4.1(75)?SSS4.2(83) 111? 95 ?SSS5.1(95)?SSS5.2(100) ?SSS5.3(109)?SSS5.4(113) ?SSS5.5(114) 111888?nnn444?AAA.125 ?SSS6.1(125)?SSS6.2(136) ?SSS6.3(142)?SSS6.4(148) 111? . 170 ?SSS7.1(170)?SSS7.2(172) ?SSS7.3(179)?SSS7.4(187) 111lll? . 194 ?SSS8.2(194)?SSS8.3(204) ?SSS8.4(210)?SSS8.5(232) 111? . 244 ?SSS9.1(?)(244)?SSS9.1(?)(252) ?SSS9.1(nnn)(266)?SSS9.2(?)(273) ?SSS9.2(?)(290)?SSS9.3(298) ?SSS9.4(309) 1 111?.323 ?SSS10.1(323)?SSS10.2(132) ?SSS10.3(334)?SSS10.4(352) 111? . 366 ?SSS11.1(366)?SSS11.2(375) ?SSS11.3(378)?SSS11.4(389) 111?222?CCC? 393 ?SSS12.1(393)?SSS12.2(400) ?SSS12.3(406) 111?nnn? . 425 ?SSS13.1(?)(425)?SSS13.1(?)(430) ?SSS13.2(493) 111?ooo?.465 2 1? ? S K7.1 (56?1 280 ?) 1. e?: (1) Z (x + 1)dx. ) Z (x + 1)dx = Z (x + 2x + 1)dx = x2 2 + 4 3x 3 2+ x + C. (3) Z 3 x2 4 x xdx. ) Z 3 x2 4 x xdx = Z (x 1 6 x 1 4)dx = 6 7x 7 6 4 3x 3 4+ C. (5) Z (2x+ 3x)2dx. ) Z (2x+ 3x)2dx = Z (4x+ 2 6 + 9x)dx = 4x ln4 + 2 ln66 x + 9x ln9 + C. (7) Z 2 3x 5 2x 3x dx. ) Z 2 3x 5 2x 3x dx = Z h 2 5 ?2 3 ?xi dx = 2x ln 5 ln2 ln3 ?2 3 ?x + C. (9) Z 1 + x + x2 x(1 + x2) dx. ) Z 1 + x + x2 x(1 + x2) dx = Z 1 + x2 x(1 + x2)dx + x x(1 + x2)dx = Z dx x + dx x (1 + x2) = ln|x| + arctgx + C. (11) Z tg2xdx. ) Z tg2xdx = Z sin2x cos2 dx = Z 1 cos2x cos2 dx = Z dx cos2 Z dx = tg x + C. 3. ?,T?LA(1,0),?z?:P(x,y)?2x 2,x R. ) ?y = f(x). fi ?f 0(x) = 2x 2,l? f(x) = Z (2x 2)dx = x2x + C. qfi ?y = f(x)LA(1,0),kf(1) = 12 2 2 + C = 0,=C = 1. u,? y = x2 2x + 1 = (x 1)2. 3 4. e?y = f(x)?:(x,y)?x3?, ?LA(1,6)?B(2,9), T?. ) fi ?x R,kf 0(x) = kx3(kX),l? f(x) = k Z x3dx = k 4x 4 + C, qfi ?LA(1,6)?B(2,9),k f(1) = k 4 + C = 6. f(2) = 4k + C = 9. l?|)?:k = 4,C = 7.u.?“y = x4+ 7. ? S K7.2 (56?1 295 ?) 1. A?e?: (1) Z xcosxdx. ) Z xcosxdx = Z xdsinx = xsinx Z sinxdx = xsinx + cosx + C. (3) Z ln(1 x)dx. ) Z ln(1 x)dx = xln(1 x) Z xdln(1 x) =xln(1 x) + Z x 1 xdx =xln(1 x) Z 1 x 1 1 x dx =xln(1 x) Z ? 1 1 1 x ? dx =xln(1 x) x ln(1 x) + C =(x 1)ln(1 x) x + C. (5) Z xnlnxdx. ) Z xnlnxdx = Z lnxd ?xn+1 n + 1 ? = xn+1 n + 1 lnx Z xn+1 n + 1dlnx = xn+1 n + 1 lnx 1 n + 1 Z xndx 4 = xn+1 n + 1 ? lnx 1 n + 1 ? + c. (7) Z excosxdx. ) ?I = Z excosxdx = Z cosxdex =excosx Z exdcosx = excosx exsinxdx =excosx + Z sinxdex =excosx + exsinx Z exdsinx =ex(cosx + sinx) excosxdx =ex(cosx + sinx) I 2I = ex(cosx + sinx),=I = Z excosxdx = 1 2e x(cosx + sinx) + C 2. A?e?: (1) Z e5xdx. ) Z e5xdx = 1 5 Z e5xd(5x) = 1 5e 5x + C. (3) Z dx 4 3x. ) Z dx 4 3x = 1 3 Z d(4 3x) 4 3x = 1 3 ln|4 3x| + C. (5) Z dx cos27x. ) Z dx cos27x = 1 7 Z d7x cos27x = 1 7tg7x + C. (7) Z cos3xsinxdx. ) Z cos3xsinxdx = Z cos3dcosx = 1 4 cos4x + C. (9) Z x2 x3 + 1dx. ) Z x2 x3 + 1dx = 1 3 Z (x3+ 1) 1 2d(x3+ 1) = 2 3 p x3+ 1 + C. (11) Z sinx cos3xdx. ) Z sinx cos3xdx = Z dcosx cos3x = 1 2cos2x + C. (13) Z dx tgx 1cos2 x. ) Z dx tgx 1cos2 x = Z d(tgx 1) tgx 1= 2ptgx 1 + C. 5 (15) Z tgx + 1 cos2x dx. ) Z tgx + 1 cos2x dx = Z ptgx + 1d(tgx + 1) =2 3(tgx + 1) 3 2+ C. (17) Z sin3x cos43xdx. ) Z sin3x 3 cos4 3xdx = 1 3 Z dcos3x 3 cos4 3x = 1 3 cos3x+ C. (19) Z arcsinx 1 x2dx. ) Z arcsinx 1 x2dx = Z arcsinxdarcsinx = 1 2(arcsinx) 2 + C. (21) Z x 1 + x2dx. ) Z x 1 + x2dx = 1 2 Z d(1 + x2) 1 + x2 = 1 2 ln(1 + x2) + C. (23) Z cosx 2sinx + 3dx. ) Z cosx 2sinx + 3dx = 1 2 = Z d(2sinx + 3) 2sinx + 3 = 1 2 ln|2sinx + 3| + C. (25) Z 2x(x2+ 1)4dx. ) Z 2x(x2+ 1)4dx = Z (x2+ 1)4d(x2+ 1) = 1 5(x 2 + 1)5+ C. (27) Z esinxcosxdx. ) Z esinxcosxdx = Z esinxdesinx= esinx+ C. (29) Z dx 1 3x2. ) Z dx 1 3x2= 1 3 Z d(3x) q 1 (3x)2 = 1 3arcsin(3x)+C. (31) Z x 1 x4dx. ) Z x 1 x4dx = 1 2 Z d(x2) 1 x4= 1 2 arcsinx2+ C. (33) Z cosxdx a2+ sin2x ) Z cosxdx a2+ sin2x = 1 a Z 1 1 + ? sinx a ?d ? sinx a ? = 1 aarctg sinx a + C. (35) Z 1 + lnx x dx = Z 1 + lnxd(1 + lnx) =2 3(1 + lnx) 3 2+ C. 3. A?e?: (1) Z arcsinxdx. 6 ) Z arcsinxdx = xarcsinx Z xdarcsinx =xarcsinx Z x 1 x2dx =xarcsinx + 1 2 Z d(1 x2) 1 x2 =xarcsinx + p 1 x2+ C. (3) Z ln(x + p 1 + x2)dx. ) Z ln(x + p 1 + x2)dx = xln(x + p 1 + x2) Z xdln(x + p 1 + x2) =xln(x + p 1 + x2) Z x x + 1 + x2 ? 1 + x 1 + x2 ? dx =xln(x + p 1 + x2) 1 2 Z d(1 + x2) 1 + x2 =xln(x + p 1 + x2) p 1 + x2+ C. (5) Z arcsin x xdx. ) Z arcsin x xdx = 2 Z arcsin xdx =2xarcsin x 2 Z xdarcsinx =2xarcsin x Z d(1 x) 1 x =2xarcsin x + 21 x + C. (7) Z xarctgxdx. ) Z xarctgxdx = 1 2 Z arctgxd(x2) = 1 2x 2arctgx 1 2 Z x2darctgx = x2 2 arctgx 1 2 Z x2 1 + x2dx = x2 2 arctgx 1 2 Z ? 1 1 1 + x2 ? dx = x2 2 arctgx x 2 + 1 2arctgx + C = 1 2(x 2 + 1)arctgx x 2 + C. 3. e?: (1) Z dx xx2 1. 7 ) Z dx xx2 1 = Z dx x|x| s 1 1 x2 = Zd ? 1 |x| ? s 1 ? ? ? 1 x ? ? ? 2 = arcsin 1 |x| + C. (3) Z 1 1 x2 ln 1 + x 1 xdx. ) Z 1 1 x2 ln 1 + x 1 xdx = 1 2 Z ln 1 + x 1 xd ? ln 1 + x 1 x ? = 1 4 ? ln 1 + x 1 x ?2 + C. (5) Z dx 1 + ex. ) Z dx 1 + ex = Z 1 + ex ex 1 + ex dx = Z dx Z ex 1 + exdx = Z dx Z d(1 + ex) 1 + ex = x ln(1 + ex) + C. (7) Z cos5xsinxdx. ) Z cos5xsinxdx = Z (1 sin2x)2sinxdsinx = Z (sin 1 2x 2sin 5 2x + sin 9 2x) = 2 3 sin 3 2x 4 7 sin 7 2x + 2 11 sin 11 2x + C. (9) Z dx 2x2 x + 2. ) Z dx 2x2 x + 2 = 1 2 Z dx s x2 x 2 + 1 = 1 2 Z d ? x 1 4 ? s 15 16 + ? x 1 4 ?2 = 1 2ln ? ? ?x 1 4 + s 15 16 + ? x 1 4 ?2? ? ? + C d7.218 = 1 2ln ? ? ?x 1 4 + p 2x2 x + 2 ? ? ? + C. 8 (11) Z (|1 + x| |1 x|)dx. ) Z (|1 + x| |1 x|)dx = Z |1 + x|d(1 + x) + Z |1 x|d(1 x) =sgn(1 + x) Z (1 + x)d(1 + x) + sgn(1 x) Z (1 x)d(1 x) =sgn(1 + x) (1 + x)2 2 + sgn(1 x) (1 x)2 2 + C = 1 2|1 + x|(1 + x) + |1 x|(1 x) + C. ? S K7.3 (56?1 305 ?) e?kn? (2) Z dx (x + 1)(x + 2)(x + 3). ) ? 1 (x + 1)(x + 2)(x + 3) = A x + 1 + B x + 2 C x + 3,k 1 = A(x + 2)(x + 3) + B(x + 1)(x + 3) + C(x + 1)(x + 2). -x = 1,k1 = 2AA = 1 2, x = 2,k1 = BB = 1, x = 3,k1 = 2CC = 1 2. l?, 1 (x + 1)(x + 2)(x + 3) = 1 2(x + 1) 1 x + 2 1 2(x + 3). u, Z dx (x + 1)(x + 2)(x + 3) = 1 2 Z dx x + 1 Z dx x + 2 + 1 2 Z dx x + 3 = 1 2 ln|x + 1| ln|x + 2| + 1 2 ln|x + 3| + C = 1 2 ln ? ? ? (x + 1)(x + 3) (x + 2)2 ? ? ? + C. (4) Z x3 1 4x3 xdx 9 ) x3 1 4x3 x = 1 4 ? 1 + x 4 4x3 x ? . ? x 4 4x3 x = x 4 x(2x 1)(2x + 1) = A x + B 2x 1 + C 2x + 1,k x 4 = A(2x 1)(2x + 1) + Bx(2x + 1) + Cx(2x 1). -x = 0,k4 = AA = 4, x = 1 2, k 7 2 = BB = 7 2, x = 1 2, k 9 2 = CC = 9 2. l?, x3 1 4x3 x = 1 4 ? 1 + 4 x 7 2(2x 1) 9 2(2x + 1) ? . u Z x3 1 4x3 xdx = 1 4 ?Z dx + 4 Z dx x 7 2 Z dx 2x 1 9 2 Z dx 2x + 1 ? = 1 4 ? x + 4ln|x| 7 4 ln|2x 1| 9 4 ln|2x + 1| ? + C = x 4 + 1 16 ln ? ? ? x16 (2x 1)7(2x + 1)9 ? ? ? + C. (6) Z dx x4+ 1. )? ? 1 x4+ 1 = 1 (x2+ 2x + 1)(x2 2x + 1)= Ax + B x2+ 2x + 1+ Cx + D x2 2x + 1. k 1 =(Ax + B)(x2 2x + 1) + (Cx + D)(x2 + 2x + 1) =(A + C)x3+ (B + D) + 2(C A)x2 + (A + C) + 2(D B)x + B + D. A + C = 0, B + D + 2(C A) = 0, A + C + 2(D B) = 0, B + D = 1. )?:A = 1 22,B = 1 2,C = 1 2. l?, 1 x4+ 1 = 1 22 h x + 2 x2+ 2x + 1 x 2 x2 2x + 1 i . u, Z dx x4+ 1 = 1 22 hZ x + 2 x2+ 2x + 1dx Z x 2 x2 2x + 1dx i 10 = 1 42 ?Z 2x 2 2 x2+ 2x + 1dx Z 2x + 2 +2 x2+ 2x + 1dx ? = 1 42 ?Z d(x2+ 2x + 1) x2+ 2x + 1+ 2Z dx ? x + 1 2 ?2 + ? 1 2 ?2 Z d(x2 2x + 1) x2 2x + 1+ 2Z dx ? x 1 2 ?2 + ? 1 2 ?2 ? = 1 42ln|x 2 + 2x + 1| + 2arctg(2x + 1) ln|x2 2x + 1| + 2arctg(2x 1) + C = 1 42 ? ln ? ? ? ? x2+ 2x + 1 x2 2x + 1 ? ? ? ?+ 2arctg 22x 1 (2x2 1) ? + C = 1 42 ? ln ? ? ? ? x2+ 2x + 1 x2 2x + 1 ? ? ? ?+ 2arctg 2x 1 x2 ? + C. )? ) k 1 x4+ 1 = 1 2 ?x2 + 1 x4+ 1 x2 1 x4+ 1 ? . u, Z dx x4+ 1 = 1 2 ?Z x2+ 1 x4+ 1dx Z x2 1 x4+ 1dx ? = 1 2 “Z 1 + 1 x2 x2+ 1 x2 dx Z 1 1 x2 x2+ 1 x2 dx # = 1 2 “Z d ? x 1 x ? ? x 1 x ?2 + 2 Z d ? x + 1 x ? ? x + 1 x ?2 2 # = 1 2 “ 1 2arctg x 1 x 2 1 22 ln ? ? ? ? ? x + 1 x 2 x + 1 x + 2 ? ? ? ? ? # + C (?7.3.4) = 1 42 ? ? ? x2+ 2x + 1 x2 2x + 1 ? ? ? + 2arctg x2 1 2x ? 5 ?)?(J,?,/“?k?,?arctg 2x 1 x2 ?arctg x2 1 2x=? . (8) Z 3x + 5 (x2+ 2x + 2)2dx. ) Z 3x + 5 (x2+ 2x + 2)2dx = Z 3x + 3 + 2 (x2+ 2x + 2)2dx 11 = 3 2 Z (2x + 2)dx (x2+ 2x + 2)2 + 2 Z dx (x2+ 2x + 2)2 eOO?z?: 1) Z (2x + 2)dx (x2+ 2x + 2)2 = Z d(x2+ 2x + 2) (x2+ 2x + 2)2 = 1 x2+ 2x + 2 + C1. 2) Z dx (x2+ 2x + 2)2 = Z dx (x + 1)2+ 12(?x + 1 = y,dx = dy) = Z dy (1 + y2)2 = Z 1 + y2 y2 (1 + y2)2 dy = Z dy (1 + y2) Z y2 (1 + y2)2dy = Z dy (1 + y2) + 1 2 Z yd ? dy 1 + y2 ? = Z dy (1 + y2) + y 2(1 + y2) 1 2 Z dy (1 + y2) = 1 2arctgy + y 2(1 + y2) + C2 = 1 2arctg(1 + x) + 1 + x 2(x2+ 2x + 2) + C2 u, Z 3x + 5 (x2+ 2x + 2)2dx = 3 2(x2+ 2x + 2) + arctg(1 + x) + 1 + x 2(x2+ 2x + 2) + C = 2x 1 2(x2+ 2x + 2) + arctg(1 + x) + C. (10) Z dx (x + 1)(x2+ x + 1)2. ) ? 1 (x + 1)(x2+ x + 1)2 = A x + 1 + Bx + C x2+ x + 1 + Dx + E (x2+ x + 1)2 k 1 = A(x2+ x + 1)2+ (Bx + C)(x + 1)(x2+ x + 1) + (Dx + E)(x + 1) )? A = 1,B = 1,C = 0,D = 1,E = 0. l? 1 (x + 1)(x2+ x + 1)2 = 1 x + 1 x x2+ x + 1 x (x2+ x + 1)2 u, Z dx (x + 1)(x2+ x + 1)2 = Z dx x + 1 Z xdx x2+ x + 1 Z xdx (x2+ x + 1)2 12 eOO?z?: 1) Z dx x + 1 = ln|x + 1| + C1. 2) Z xdx x2+ x + 1 = 1 2 Z 2x + 1 1 x2+ x + 1dx = 1 2 “Z d(x2+ x + 1) x2+ x + 1 Z d ? x + 1 2 ? ? x + 1 2 ?2 + ?3 2 ?2 # = 1 2 ln|x2+ x + 1| 1 3arctg 2x + 1 3+ C2. 3) Z xdx (x2+ x + 1)2 = 1 2 Z 2x + 1 1 (x2+ x + 1)2dx = 1 2 ?Z d(x2+ x + 1) (x2+ x + 1)2 Z dx (x2+ x + 1)2 ? , ? Z d(x2+ x + 1) (x2+ x + 1)2 = dx x2+ x + 1 + C 0 Z dx (x2+ x + 1)2 = Z d ? x + 1 2 ? h? x + 1 2 ?2 + ?3 2 ?2i (d4?“) = x + 1 2 2 3 4 h? x + 1 2 ?2 + ?3 2 ?2i + 1 2 3 4 Z d ? x + 1 2 ? ? x + 1 2 ?2 + ?3 2 ?2 = 2x + 1 3(x2+ x + 1) + 4 33arctg 2x + 1 3+ C 00. l?, Z xdx (x2+ x + 1)2 = 1 2(x2+ x + 1) 2x + 1 6(x2+ x + 1) 13 2 33arctg 2x + 1 3+ C3 = x + 2 3(x2+ x + 1) 2 33arctg 2x + 1 3+ C3 u, Z dx (x + 1)(x2+ x + 1)2 =ln|x + 1| 1 2 ln|x2+ x + 1| + 1 3arctg 2x + 1 3 + x + 2 3(x2+ x + 1) + 2 33arctg 2x + 1 3+ C =ln|x + 1| + 1 2 ln|x2+ x + 1| + x + 2 3(x2+ x + 1) + 5 33arctg 2x + 1 3+ C. ? S K7.4 (56?1 316 ?) 1. e? (1) Z x 1 3 x + 1dx. ) ?t = 6 x x = t6,dx = 6t5dt.k Z x 1 3 x + 1dx = 6 Z t3 1 t2+ 1 t5dt = Z ? t6 t4 t3+ t2+ t + 1 t t2+ 1 + 1 t2+ 1 ? dt =6 ?t7 7 t5 5 t4 4 + t3 3 + t2 2 t + 1 2 ln(t2+ 1) + arctgt ? + C = 6 7x 7 6 6 5x 5 6 3 2x 2 3+ 2x 1 2+ 3x 1 3 6x 1 6 3ln|xfrac13+ 1| + 6arctgx 1 6+ C. (3) Z 2 + x 3 3 xdx. ) ?t = 3 3 x x = 3 t3,dx = 3t2dt.k Z 2 + x 3 3 xdx = Z 2 + (3 t3) t (3t2)dt = Z (t4 5t)dt 14 =3 ?1 5t 5 5 2t 2? + C = 3t2 ?1 5t 3 5 2 ? + C =3(3 x) 2 3 ?1 5(3 x) 5 2 ? + C = 3 5(3 x) 2 3 ? x + 19 2 ? + C. (5) Z s 2 + 3x x 3 dx. ) ?t = s 2 + 3x x 3 x = 3t2+ 2 t2 3 ,dx = 22t (t2 3)2dt.k Z s 2 + 3x x 3 dx = Z t 22t (t2 3)2dt = 22 Z t2 (t2 3)3dt =11 Z td ? 1 t2 3 ? = 11 ? t t2 3 Z dt t2 3 ? =11 h t t2 3 1 23 ln ? ? ? t 3 t + 3 ? ? ? i + C 0 = p 3x2 7x 6 11 23 ln 11 6 ? ? ? ? ?x 7 6 + s x2 7 3x 2 ? ? ? ? ? + C 0 = p 3x2 7x 6 + 11 23 ln ? ? ? ? ?x 7 6 + s x2 7 3x 2 ? ? ? ? ? + C. (7) , Z dx 2x x2. ) Z dx 2x x2= Z d(x 1) p1 (x 1)2= arcsin(x 1) + C. (9) Z x + 3 1 4x2dx. ) Z x + 3 1 4x2dx = 1 8 Z d(1 4x2) 1 4x2+ 3 2 Z d(2x) p1 (2x)2 = 1 4 p 1 4x2+ 3 2 arcsin2x + C. (11) Z 3x + 5 px(2x 1)dx. 15 ) Z 3x + 5 px(2x 1)dx = Z 3x 3 4 + 3 4 + 5 px(2x 1)dx = 3 4 Z 4x + 1 2x2 xdx + 23 4 Z dx 2x2 x = 3 4 Z d(2x2 x) 2x2 x + 23 42 Z d2 ? x 1 4 ? sh 2?x 1 4 ?i2 1 8 = 3 2 p 2x2 x + 23 42 ln 2 4 |4x 1 + 8x(2x 1)| + C0 = 3 2 p 2x2 x + 23 42 ln|4x 1 + 8x(2x 1)| + C. (13) Z dx x x2 1. ) Z dx x x2 1 = Z (x + x2 1) (x x2 1)(x + x2 1)dx = Z (x + p x2 1)dx = Z xdx + Z p x2 1 dx = x2 2 + x 2 p x2 1 1 2 ln|x + p x2 1| + C. (15) Z x + 1 (2x + x2)2x + x2dx. ) Z x + 1 (2x + x2)2x + x2dx = 1 2 Z d(2x + x2) (2x + x2) 1 2 = 1 2x + x2+ C. 3. e?: (1) Z cos4xsin3xdx. ) Z cos4xsin3xdx = Z cos4xsin2xsinxdx = Z cos4x(1 cos2x)dcosx = Z (cos6cos4x)dcosx = 1 7 cos7x = 1 5 cos5x + C. (3) Z sin4xcos4xdx. ) Z sin4xcos4xdx = Z (sinxcosx)4= Z ?1 2 sin2x ?4 dx 16 = 1 16 Z (sin22x)2dx = 1 16 Z ?1 cos4x 2 ?2 dx = 1 64 Z (1 2cos4x + cos24x)dx = 1 64 Z h (1 2cos4x + 1 2(1 + cos8x) i dx = 1 64 Z ?3 2 2cos4x + 1 2 cos8x ? dx = 1 64 ?3 2x 1 2 sin4x + 1 16 sin8x ? + C = 1 64 ? 3x sin4x + 1 8 sin8x ? + C. (5) Z ctg3xdx. ) Z ctg3xdx = Z ctg2xctgxdx = Z (1 csc2x)sinxdcscx = Z ? 1 cscx cscx ? dcscx =ln|cscx| 1 2 csc2x + C 0 = ln|sinx| 1 2ctg 2x + C. (7) Z sec8xdx. ) ?t = tgx,dt = secxdx.k Z sec8