1、習(xí)題 41 1. 求下列不定積分: (1)dx x2 1 ; 解 C x Cxdxxdx x +=+ + = + 1 12 11 122 2 . (2)dxxx; 解 CxxCxdxxdxxx+=+ + = + 2 1 2 3 2 3 5 2 1 2 3 1 . (3)dx x 1 ; 解 CxCxdxxdx x +=+ + = + 2 1 2 1 11 1 2 1 2 1 . (4)dxxx 32 ; 解 CxxCxdxxdxxx+=+ + = + 33 1 3 7 3 7 32 10 3 1 3 7 1 . (5)dx xx2 1 ; 解 C xx Cxdxxdx xx +=+ + = +

2、 1 2 3 1 2 5 11 1 2 5 2 5 2 . (6)dxx mn ; 解 Cx mn m Cx m n dxxdxx m nm m n m n mn + + =+ + = + + 1 1 1 . (7); dxx35 解 Cxdxxdxx+= 433 4 5 55. (8); +dxxx)23( 2 解 Cxxxdxdxxdxxdxxx+=+=+ 2 2 3 3 1 23)23( 2322 . (9) gh dh 2 (g 是常數(shù)); 解 C g h Ch g dhh ggh dh +=+= 2 2 2 1 2 1 2 2 1 2 1 . (10); dxx 2 )2( 解 Cx

3、xxdxdxxdxxdxxxdxx+=+=+= 42 3 1 44)44()2( 23222 . (11); +dxx 22 ) 1( 解 Cxxxdxdxxdxxdxxxdxx+=+=+=+ 35242422 3 2 5 1 2) 12() 1(. (12)dxxx +) 1)(1( 3 ; 解 +=+=+dxdxxdxxdxxdxxxxdxxx 2 3 2 1 2323 ) 1() 1)(1( Cxxxx+= 2 5 2 3 3 5 2 3 2 3 1 . (13) dx x x 2 )1 ( ; 解 Cxxxdxxxxdx x xx dx x x +=+= + = 2 5 2 3 2 1

4、 2 3 2 1 2 1 22 5 2 3 4 2)2( 21)1 ( . (14) + + dx x xx 1 133 2 24 ; 解 Cxxdx x xdx x xx += + += + + arctan) 1 1 3( 1 133 3 2 2 2 24 . (15) + dx x x 2 2 1 ; 解 += + = + + = + Cxxdx x dx x x dx x x arctan) 1 1 1 ( 1 11 1 22 2 2 2 . (16)+dx x ex) 3 2(; 解 Cxedx x dxedx x e xxx +=+=+ |ln32 1 32) 3 2(. (17)

5、 + dx x x ) 1 2 1 3 ( 2 2 ; 解 += + = + Cxxdx x dx x dx x x arcsin2arctan3 1 1 2 1 1 3) 1 2 1 3 ( 2 2 2 2 . (18)dx x e e x x )1 (; 解 Cxedxxedx x e e xx x x += 2 1 2 1 2)()1 (. (19); dxex x 3 解 C e C e e dxedxe xxx xxx + + =+= 13ln 3 )3ln( )3( )3(3. (20) dx x xx 3 2532 ; 解 CxCxdxdx x x x x xx + =+= )

6、3 2 ( 3ln2ln 5 2 3 2 ln ) 3 2 ( 52) 3 2 (52 3 2532 . (21); dxxxx)tan(secsec 解 . +=Cxxdxxxxdxxxxsectan)tansec(sec)tan(secsec 2 (22)dx x 2 cos2; 解 Cxxdxxdx x dx x +=+= + = )sin( 2 1 )cos1 ( 2 1 2 cos1 2 cos2. (23) + dx x2cos1 1 ; 解 += + Cxdx x dx x tan 2 1 cos2 1 2cos1 1 2 . (24) dx xx x sincos 2cos ;

7、 解 +=+= = Cxxdxxxdx xx xx dx xx x cossin)sin(cos sincos sincos sincos 2cos 22 . (25)dx xx x 22 sincos 2cos ; 解 += =Cxxdx xx dx xx xx dx xx x tancot) cos 1 sin 1 ( sincos sincos sincos 2cos 2222 22 22 . (26)dxxx x ) 1 1 ( 2 ; 解 dxxx x2 1 1 += Cxxdxxx 4 1 4 7 4 5 4 3 4 7 4 )(. 2. 一曲線通過點(e2, 3), 且在任一點處

8、的切線的斜率等于該點橫坐標(biāo)的倒數(shù), 求該曲 線的方程. 解 設(shè)該曲線的方程為 y=f(x), 則由題意得 x xfy 1 )(=, 所以 Cxdx x y+=|ln 1 . 又因為曲線通過點(e2, 3), 所以有=32=1 3=f(e 2)=ln|e 2|+C=2+C, C=32=1. 于是所求曲線的方程為 y=ln|x|+1. 3. 一物體由靜止開始運動, 經(jīng)t秒后的速度是 3t2(m/s), 問 (1)在 3 秒后物體離開出發(fā)點的距離是多少？ (2)物體走完 360m 需要多少時間？ 解 設(shè)位移函數(shù)為s=s(t), 則s=v=3 t2, . Ctdtts+= 32 3 因為當(dāng)t=0 時,

9、 s=0, 所以C=0. 因此位移函數(shù)為s=t 3. (1)在 3 秒后物體離開出發(fā)點的距離是s=s(3)=33=27. (2)由t 3=360, 得物體走完 360m所需的時間11. 7360 3 =ts. 4. 證明函數(shù) x e2 2 1 , exshx和exchx都是 xx e x shch 的原函數(shù). 證明 x x x xxxx xx e e e eeee e xx e 2 22 shch = + = . 因為 xx ee 22 ) 2 1 (=, 所以 x e2 2 1 是 xx e x shch 的原函數(shù). 因為 (exshx)=exshx+exchx=ex(sh x+ch x)

10、x xxxx x e eeee e 2 ) 22 (= + + = , 所以exshx是 xx e x shch 的原函數(shù). 因為 (exchx)=exchx+exshx=ex(ch x+sh x) x xxxx x e eeee e 2 ) 22 (= + + = , 所以exchx是 xx e x shch 的原函數(shù). 習(xí)題 42 1. 在下列各式等號右端的空白處填入適當(dāng)?shù)南禂?shù), 使等式成立(例如: )74( 4 1 +=xddx: (1) dx= d(ax); 解 dx= a 1 d(ax). (2) dx= d(7x3); 解 dx= 7 1 d(7x3). (3) xdx= d(x2

11、); 解xdx= 2 1 d(x2). (4) xdx= d(5x2); 解xdx= 10 1 d(5x2). (5); )1 ( 2 xdxdx= 解 )1 ( 2 1 2 xdxdx=. (6)x3dx= d(3x42); 解x3dx= 12 1 d(3x42). (7)e 2x dx= d(e2x); 解e 2x dx= 2 1 d(e2x). (8)1 ( 22 xx eddxe +=; 解 )1 ( 2 22 xx eddxe +=. (9) 2 3 (cos 2 3 sinxdxdx=; 解 ) 2 3 (cos 3 2 2 3 sinxdxdx=. (10)|)|ln5( xd

12、x dx =; 解 |)|ln5( 5 1 xd x dx =. (11)|)|ln53( xd x dx =; 解 |)|ln53( 5 1 xd x dx =. (12)3(arctan 91 2 xd x dx = + ; 解 )3(arctan 3 1 91 2 xd x dx = + . (13)arctan1 ( 1 2 xd x dx = ; 解 )arctan1 ( ) 1( 1 2 xd x dx = . (14)1( 1 2 2 xd x xdx = . 解 )1( ) 1( 1 2 2 xd x xdx = . 2. 求下列不定積分(其中 a, b, , 均為常數(shù)): (

13、1); dte t 5 解 Cexdedte xxt += 555 5 1 5 5 1 . (2); dxx 3 )23( 解 Cxxdxdxx+= 433 )23( 8 1 )23()23( 2 1 )23(. (3) dx x21 1 ; 解 Cxxd x dx x += = |21 |ln 2 1 )21 ( 21 1 2 1 21 1 . (4) 3 32x dx ; 解 CxCxxdx x dx +=+= 3 2 3 2 3 1 3 )32( 2 1 )32( 2 3 3 1 )32()32( 3 1 32 . (5)dxeax b x )(sin; 解 Cbeax ab x deb

14、axdax a dxeax b x b x b x += cos 1 )()(sin 1 )(sin. (6)dt t tsin ; 解 +=Cttdtdt t t cos2sin2 sin . (7); xdxx 210 sectan 解 xdxx 210 sectanCxxxd+= 1110 tan 11 1 tantan. (8) xxx dx lnlnln ; 解 Cxxd x xd xxxxx dx += |lnln|lnlnln lnln 1 ln lnlnln 1 lnlnln . (9) + +dx x x x 2 2 1 1tan; 解 + +dx x x x 2 2 1 1

15、tan 2 2 2 22 1 1cos 1sin 11tanxd x x xdx+ + + =+= Cxxd x +=+ + = |1cos|ln1cos 1cos 1 22 2 . (10) xx dx cossin ; 解 Cxxd x dx x x xx dx += |tan|lntan tan 1 tan sec cossin 2 . (11) + dx ee xx 1 ; 解 + dx ee xx 1 Cede e dx e e xx xx x += + = + = arctan 1 1 1 22 . (12); dxxe x2 解 . 2 1 )( 2 1 222 2 Cexded

16、xxe xxx += (13); dxxx)cos( 2 解 Cxxdxdxxx+= )sin( 2 1 )()cos( 2 1 )cos( 2222 . (14) dx x x 2 32 ; 解 CxCxxdxdx x x +=+= 2 2 1 22 2 1 2 2 32 3 1 )32( 3 1 )32()32( 6 1 32 . (15) dx x x 4 3 1 3 ; 解 += = Cxxd x dx x x |1 |ln 4 3 )1 ( 1 1 4 3 1 3 44 44 3 . (16); +dttt)sin(cos2 解 Cttdtdttt+=+=+ )(cos 3 1 )c

17、os()(cos 1 )sin()(cos 322 . (17)dx x x 3 cos sin ; 解 CxCxxxddx x x +=+= 223 3 sec 2 1 cos 2 1 coscos cos sin . (18) + dx xx xx 3 cossin cossin ; 解 )sincos( cossin 1 cossin cossin 33 xxd xx dx xx xx + = + Cxxxxdxx+= 3 2 3 1 )cos(sin 2 3 )cos(sin)cos(sin. (19) dx x x 2 49 1 ; 解 dx x x dx x dx x x = 22

18、2 4949 1 49 1 )49( 49 1 8 1 ) 3 2 ( ) 3 2 (1 1 2 1 2 2 2 xd x xd x + = Cx x += 2 49 4 1 3 2 arcsin 2 1 . (20) + dx x x 2 3 9 ; 解 Cxxxd x xd x x dx x x += + = + = + )9ln(9 2 1 )() 9 9 1 ( 2 1 )( 9 2 1 9 222 2 2 2 2 2 3 . (21) dx x12 1 2 ; 解 + = + = dx xx dx xx dx x ) 12 1 12 1 ( 2 1 ) 12)(12( 1 12 1

19、2 + + =) 12( 12 1 22 1 ) 12( 12 1 22 1 xd x xd x C x x Cxx+ + =+=| 12 12 |ln 22 1 | 12|ln 22 1 | 12|ln 22 1 . (22) + dx xx)2)(1( 1 ; 解 C x x Cxxdx xx dx xx + + =+= + = + | 1 2 |ln 3 1 | 1|ln| 2|(ln 3 1 ) 1 1 2 1 ( 3 1 )2)(1( 1 . (23); xdx 3 cos 解 Cxxxdxxdxxdx+= 3223 sin 3 1 sinsin)sin1 (sincoscos. (

20、24); +dtt)(cos2 解 Cttdttdtt+=+=+ )(2sin 4 1 2 1 )(2cos1 2 1 )(cos2 . (25); xdxx3cos2sin 解 xdxx3cos2sinCxxdxxx+= cos 2 1 5cos 10 1 )sin5(sin 2 1 . (26)dx x x 2 coscos; 解 Cxxdxxxdx x x+=+= 2 1 sin 2 3 sin 3 1 ) 2 1 cos 2 3 (cos 2 1 2 coscos. (27); xdxx7sin5sin 解 Cxxdxxxxdxx+= 2sin 4 1 12sin 24 1 )2cos

21、12(cos 2 1 7sin5sin. (28); xdxxsectan3 解 xdxxdxxxxdxxsectantansectansectan 223 = Cxxxdx+=secsec 3 1 sec) 1(sec 32 . (29) dx x x 2 arccos2 1 10 ; 解 Cxdxddx x x xx x += 10ln2 10 )arccos2(10 2 1 arccos10 1 10 arccos2 arccos2arccos2 2 arccos2 . (30) + dx xx x )1 ( arctan ; 解 Cxxdxxd x x dx xx x += + = +

22、 2 )(arctanarctanarctan2 )1 ( arctan 2 )1 ( arctan . (31) 22 1)(arcsinxx dx ; 解 C x xd x xx dx += arcsin 1 arcsin )(arcsin 1 1)(arcsin 2 22 . (32) + dx xx x 2 )ln( ln1 ; 解 C xx xxd xx dx xx x += + ln 1 )ln( )ln( 1 )ln( ln1 22 . (33)dx xx x sincos tanln ; 解 =xd x x xdx x x dx xx x tan tan tanln sec t

23、an tanln sincos tanln 2 Cxxdx+= 2 )tan(ln 2 1 tanlntanln. (34) dx xa x 22 2 (0); a 解 = = dt t adttatdta ta tatax dx xa x 2 2cos1 sincos cos sinsin 222 22 22 2 令 , Cxa x a xa Ct a ta+=+= 22 22 2 2 arcsin 2 2sin 42 1 . (35) 1 2 xx dx ; 解 C x Ctdttdtt tt tx xx dx +=+= = 1 arccostansec tansec 1sec 1 2 令

24、 . 或 C xx d x dx x x xx dx += = = 1 arccos 1 1 1 1 1 1 1 1 22 2 2 . (36) + 32 ) 1(x dx ; 解 Cttdttd t tx x dx += + = + sincostan ) 1(tan 1tan ) 1( 3232 令 C x x + + = 1 2 . (37) dx x x9 2 ; 解 = = tdttd t ttx dx x x 2 22 tan3)sec3( sec3 9sec9sec39令 C x xCttdt t +=+= 3 arccos393tan3) 1 cos 1 (3 2 2 . (3

25、8 ) +x dx 21 ; 解 CxxCttdt t tdt t tx x dx +=+= + = + = + )21ln(2)1ln() 1 1 1 ( 1 12 21 令 . (39) + 2 11x dx ; = + = + = + dt t dt t tdt t tx x dx ) 2 sec 2 1 1 () cos1 1 1 (cos cos1 1sin 11 2 2 令 解 C x x xC t t tC t t+ + =+ + =+= 2 11 arcsin cos1 sin 2 tan . (40) + 2 1 xx dx . 解 + + = + = + dt tt ttt

26、t tdt tt tx xx dx cossin sincossincos 2 1 cos cossin 1sin 1 2 令 Ctttttd tt dt+=+ + += |cossin|ln 2 1 2 1 )cos(sin cossin 1 2 1 2 1 Cxxx+=|1|ln 2 1 arcsin 2 1 2 . 習(xí)題 43 求下列不定積分: 1. ; xdxxsin 解 . Cxxxxdxxxxxdxdxx+= sincoscoscoscossin 2. ; xdxln 解 . Cxxxdxxxxxdxxxdx+= lnlnlnlnln 3. ; xdxarcsin 解 =xxdxx

27、xdxarcsinarcsinarcsin =dx x x xx 2 1 arcsin Cxxx+= 2 1arcsin. 4. ; dxxe x 解 +=dxexexdedxxe xxxx CxeCexe xxx +=+= ) 1(. 5. ; xdxx ln 2 解 =xdxxxxdxxdxxln 3 1 ln 3 1 ln 3 1 ln 3332 Cxxxdxxxx+= 3323 9 1 ln 3 1 3 1 ln 3 1 . 6. ; xdxe x cos 解 因為 +=xdxexexdexexdexdxe xxxxxx sinsinsinsinsincos += xxxxx xdex

28、exexdexecoscossincossin , =xdxexexe xxx coscossin 所以 CxxeCxexexdxe xxxx +=+= )cos(sin 2 1 )cossin( 2 1 cos. 7. dx x e x 2 sin 2 ; 解 因為 = xxxx de xx e x dedx x e 2222 2 cos2 2 cos2 2 cos2 2 sin +=+= 2 sin8 2 cos2 2 cos4 2 cos2 2222 x de x edx x e x e xxxx += xxx de xx e x e 222 2 sin8 2 sin8 2 cos2 +

29、=dx x e x e x e xxx 2 sin16 2 sin8 2 cos2 222 , 所以 C xx edx x e xx += ) 2 sin4 2 (cos 17 2 2 sin 22 . 8. dx x x 2 cos; 解 C xx xdx xx x x xddx x x+= 2 cos4 2 sin2 2 sin2 2 sin2 2 sin2 2 cos. 9. ; xdxx arctan 2 解 + =dx x xxxxdxxdxx 2 3332 1 1 3 1 arctan 3 1 arctan 3 1 arctan + = + = 2 2 32 2 2 3 ) 1 1

30、 1 ( 6 1 arctan 3 1 1 6 1 arctan 3 1 dx x xxdx x x xx Cxxxx+=)1ln( 6 1 6 1 arctan 3 1 223 . 10. xdxx 2 tan 解 +=xxdxxdxxdxxdxxxxdxxtan 2 1 sec) 1(sectan 2222 Cxxxxxdxxxx+=+= |cos|lntan 2 1 tantan 2 1 22 . 11. ; xdxx cos 2 解 +=xxdxxxdxxxxxdxxdxxcos2sin2sinsinsincos 2222 . Cxxxxxxdxxxxx+=+= sin2cos2sin

31、cos2cos2sin 22 12. ; dtte t2 解 +=dtetetdedtte tttt2222 2 1 2 1 2 1 CteCete ttt +=+= ) 2 1 ( 2 1 4 1 2 1 222 . 13. ; xdx 2 ln 解 =xdxxxdx x xxxxxdxln2ln 1 ln2lnln 222 Cxxxxxdx x xxxxx+=+= 2ln2ln 1 2ln2ln 22 . 14. ; xdxxxcossin 解 +=xdxxxxxdxdxxxdxxx2cos 4 1 2cos 4 1 2cos 4 1 2sin 2 1 cossin Cxxx+=2sin

32、8 1 2cos 4 1 . 15. dx x x 2 cos2 2 ; 解 +=+=+=xdxxxxxxdxxdxxxdx x xsinsin 2 1 6 1 sin 2 1 6 1 )cos1 ( 2 1 2 cos 2323222 +=+=xdxxxxxxxxdxxxcoscossin 2 1 6 1 cossin 2 1 6 1 2323 Cxxxxxx+=sincossin 2 1 6 1 23 . 16. ; dxxx) 1ln( 解 =dx x xxxdxxdxxx 1 1 2 1 ) 1ln( 2 1 ) 1ln( 2 1 ) 1ln( 222 +=dx x xxx) 1 1

33、1( 2 1 ) 1ln( 2 1 2 Cxxxxx+=) 1ln( 2 1 2 1 4 1 ) 1ln( 2 1 22 . 17. ; xdxx2sin) 1( 2 解 +=xdxxxxxdxxdxx22cos 2 1 2cos) 1( 2 1 2cos) 1( 2 1 2sin) 1( 222 +=xxdxx2sin 2 1 2cos) 1( 2 1 2 +=xdxxxxx2sin 2 1 2sin 2 1 2cos) 1( 2 1 2 Cxxxxx+=2cos 4 1 2sin 2 1 2cos) 1( 2 1 2 . 18. dx x x 2 3 ln ; 解 +=+=xdx x x

34、x xd x x xx xddx x x 2 2 3333 2 3 ln 1 3ln 1 ln 1 ln 11 ln ln +=xd x x x x xx xdx x 22323 ln 1 3ln 3 ln 11 ln3ln 1 =+= x xdx x x x dxx x x x x x 1 ln6ln 3 ln 1 ln 1 6ln 3 ln 1 23 2 23 +=dx x x x x x x x 2 23 1 6ln 6 ln 3 ln 1 C x x x x x x x += 6 ln 6 ln 3 ln 1 23 . 19. dxe x 3 ; 解 = = ttx detdtet t

35、x dxe 22 3 33 3令 = tttt tdeetdtteet6363 22 +=dteteet ttt 663 2 Ceteet ttt +=663 2 Cxxe x +=)22(3 3 32 3 . 20. ; xdxlncos 解 因為 +=dx x xxxxxdx 1 lnsinlncoslncos dx x xxxxxxxdxxx 1 lncoslnsinlncoslnsinlncos+=+= , +=xdxxxxxlncoslnsinlncos 所以 Cxx x xdx+= )lnsinln(cos 2 lncos. 21. ; dxx 2 )(arcsin 解 =dx x

36、 xxxxdxx 2 22 1 1 arcsin2)(arcsin)(arcsin += 22 1arcsin2)(arcsinxxdxx +=dxxxxx2arcsin12)(arcsin 22 Cxxxxx+=2arcsin12)(arcsin 22 . 22. . xdxex 2 sin 解 =xdxeedxxexdxe xxxx 2cos 2 1 2 1 )2cos1 ( 2 1 sin2, 而 dxxexexdexdxe xxxx +=2sin22cos2cos2cos , +=+=xdxexexedexxe xxxxx 2cos42sin22cos2sin22cos Cxxexdx

37、e xx += )2sin22(cos 5 1 2cos, 所以 Cxxeexdxe xxx += )2sin22(cos 10 1 2 1 sin2. 習(xí)題 44 求下列不定積分: 1. dx x x +3 3 ; 解 dx x xxx dx x x dx x x + + = + + = +3 27)93)(3( 3 2727 3 233 + +=dx x dxxx 3 1 27)93( 2 Cxxxx+=| 3|ln279 2 3 3 1 23 . 2. + + dx xx x 103 32 2 ; 解 Cxxxxd xx dx xx x +=+ + = + + |103|ln)103(

38、103 1 103 32 22 22 . 3. + dx xx xx 3 45 8 ; 解 + += + dx xx xx dxxxdx xx xx 3 2 2 3 45 8 ) 1( 8 + +=dx x dx x dx x xxx 1 3 1 48 2 1 3 1 23 Cxxxxxx+=| 1|ln3| 1|ln4|ln8 2 1 3 1 23 . 4. + dx x1 3 3 ; 解 + + + + = + + + + = + dx xxxx x x dx xx x x dx x ) 1 1 2 3 1 12 2 1 1 1 () 1 2 1 1 ( 1 3 2223 + + + +=

39、) 2 1 ( ) 2 3 () 2 1 ( 1 2 3 ) 1( 1 1 2 1 | 1|ln 22 2 2 xd x xxd xx x C x xx x + + + + = 3 12 arctan3 1 | 1| ln 2 . 5. +)3)(2)(1(xxx xdx ; 解 dx xxxxxx xdx ) 3 3 1 1 2 4 ( 2 1 )3)(2)(1(+ + + = + Cxxx+=|)1|ln| 3|ln3| 2|(ln 2 1 . 6. + + dx xx x ) 1() 1( 1 2 2 ; 解 + + + = + + dx xxx dx xx x ) 1( 1 1 1 2

40、 1 1 1 2 1 ) 1() 1( 1 22 2 C x xx+ + += 1 1 | 1|ln 2 1 | 1|ln 2 1 C x x+ + += 1 1 | 1|ln 2 1 2 . 7. dx xx) 1( 1 2 + ; 解 Cxxdx x x x dx xx += + = + )1ln( 2 1 |ln) 1 1 ( ) 1( 1 2 22 . 8. +)(1( 22 xxx dx ; 解 + + + = + dx xx x xxxx dx ) 1 1 2 1 1 1 2 11 ( )(1( 222 + + +=dx x x xx 1 1 2 1 | 1|ln 2 1 |ln

41、2 + + +=dx x dx x x xx 1 1 2 1 1 2 4 1 | 1|ln 2 1 |ln 22 Cxxxx+=arctan 2 1 ) 1ln( 4 1 | 1|ln 2 1 |ln 2 . 9. +) 1)(1( 22 xxx dx ; 解 dx x x xx x xxx dx ) 11 1 ( ) 1)(1( 2222 + + + = + ) 1ln( 2 1 1 1 2 1 1 12 2 1 2 22 + + + + + = xdx xxxx x + +=dx xx xxx 1 1 2 1 ) 1ln( 2 1 | 1|ln 2 1 2 22 C x xxx+ + +=

42、 3 12 arctan 3 3 ) 1ln( 2 1 | 1|ln 2 1 22 . 10. + dx x1 1 4 ; 解 dx xxxx dx x + = +) 12)(12( 1 1 1 224 + + + + + =dx xx x dx xx x 12 2 1 4 2 12 2 1 4 2 22 + + + =dx xx x dx xx x 12 2 2 )22( 2 1 4 2 12 2 2 )22( 2 1 4 2 22 ) 1212 ( 4 1 12 ) 12( 12 ) 12( 8 2 222 2 2 2 + + + + + + + + = xx dx xx dx xx xx

43、d xx xxd Cxx xx xx + + + =) 12arctan( 4 2 ) 12arctan( 4 2 | 12 12 |ln 8 2 2 2 . 11. + dx xx x 22 2 ) 1( 2 ; 解 + + = + dx xx dx xx x dx xx x 1 1 ) 1( 1 ) 1( 2 22222 2 + + + + =dx xx dx xx dx xx x 1 1 ) 1( 1 2 3 ) 1( 12 2 1 22222 + + + =dx xx dx xxxx1 1 ) 1( 1 2 3 1 1 2 1 2222 , 因為 ) 3 12 arctan( 3 2

44、) 3 12 ( ) 3 12 (1 1 3 2 1 1 2 2 + = + + + = + xx d x dx xx , 而 + = + dx x dx xx 222 22 ) 2 3 () 2 1 ( 1 ) 1( 1 由遞推公式 + + + = + )( )32( )( ) 1(2 1 )( 122122222nnn ax dx n ax x naax dx , 得 + = + dx x dx xx 222 22 ) 2 3 () 2 1 ( 1 ) 1( 1 3 12 arctan 3 2 3 2 1 12 3 1 ) 11 2 1 ( ) 2 3 (2 1 222 2 + + + +

45、 = + + + + = x xx x xx dx xx x , 所以 + dx xx x 22 2 ) 1( 2 C xx xx x xx + + + + + + = 3 12 arctan 3 2 3 12 arctan 3 2 1 12 2 1 1 1 2 1 22 C x xx x + + + + = 3 12 arctan 3 4 1 1 2 . 12. +x dx 2 sin3 ; 解 + = = + xd x dx xx dx tan 3tan4 1 cos4 1 sin3 222 C x xd x += + = 3 tan2 arctan 32 1 tan ) 2 3 (tan

46、 1 4 1 22 . 13. + dx xcos3 1 ; 解 + = + = + ) 2 sec1 ( 2 cos ) 2 ( 2 cos1 2 1 cos3 1 222 xx x d x dx dx x += + =C x x x d 2 2 tan arctan 2 1 2 tan2 2 tan 2 . 或 + + + = + du u u u x u dx x 2 2 1 2 1 2 3 12 tan cos3 1 令 C x C u du u +=+= + = 2 2 tan arctan 2 1 2 arctan 2 1 )2( 1 22 . 14. + dx xsin2 1 ;

47、 解 + = + = + ) 2 cot 2 (csc 2 sin ) 2 ( 2 cos 2 sin22 sin2 1 22 xxx x d xx dx dx x + + = + = 22 2 ) 2 3 () 2 1 2 (cot ) 2 1 2 (cot 1 2 cot 2 cot ) 2 (cot x x d xx x d C x + + = 3 1 2 cot2 arctan 3 2 . 或 + + + = + du u u u x u dx x 2 2 1 2 1 2 2 12 tan sin2 1 令 + = + =du u du uu 22 2 ) 2 3 () 2 1 ( 1

48、 1 1 C x C u + + =+ + = 3 1 2 tan2 arctan 3 2 3 12 arctan 3 2 . 15. +xx dx cossin1 ; 解 += + = + = + C x x x d xx dx xx dx | 2 tan|ln 2 tan1 ) 2 (tan ) 2 tan1 ( 2 cos 2 1 cossin1 2 . 或 + + + + + = + du u u u u u x u xx dx 2 2 2 2 1 2 1 1 1 2 1 12 tan cossin1 令 C x Cudu u +=+= + =| 1 2 tan|ln| 1|ln 1 1 . 16. +5cossin2xx dx ; 解 + = + + + + = + du uu du u u u u u x u xx dx 223 1 1 2 5 1 1 1 4 12 tan 5cossin2 22 2 2 2 令 C x C u du u + + =+ + = + = 5 1 2 tan3 arctan 5 1 5 13 arctan 5 1 ) 3 5 () 3 1 ( 1 3 1 22 . 或 + + + + = + du u u u u u x u xx d