第八章 不定积分总练习题求下列不定积分： (1)∫43x1x 2x --dx ；(2)∫xarcsinxdx ；(3)∫x1dx +；(4)∫e sinx sin2xdx ；(5)∫xe dx ；(6)∫1x x dx2-；(7)∫x tan 1x tan 1+-dx ；(8)∫32)2-x (x-x dx ； (9)∫x cos dx 4；(10)∫sin 4xdx ；(11)∫4x 3x 5-x 23+-dx ；(12)∫arctan(1+x )dx ； (13)∫2x x 47+dx ；(14)∫x tan tanx 1tanx 2++dx ；(15)∫1002x)-(1x dx ； (16)∫2x arcsinx dx ；(17)∫xln ⎪⎭⎫ ⎝⎛+x -1x 1dx ；(18)∫xsinx cos dx 7；(19)∫e x 22x 1x -1⎪⎭⎫ ⎝⎛+dx ； (20)I n =∫uv n dx, 其中u=a 1+b 1x ，v=a 2+b 2x ，求递推形式解.解：(1)∫43x 1x 2x --dx=∫41x dx-2∫121x dx-∫41x-dx =5445x -13241213x -34∫43x +C.(2)∫xarcsinxdx=-21∫arcsinxd(1-x 2)=-21(1-x 2)arcsinx+21∫(1-x 2)darcsinx=-21(1-x 2)arcsinx+21∫2x -1dx =-21(1-x 2)arcsinx+21∫t sin -12dsint=-21(1-x 2)arcsinx+21∫cos 2tdt=-21(1-x 2)arcsinx+81∫(1+cos2t)d2t =-21(1-x 2)arcsinx+4t +81sin2t+C=-21(1-x 2)arcsinx+41arcsinx +41sintcost+C =2x 2arcsinx-41arcsinx +2x -14x+C. (3)∫x 1dx+=∫t 1dt 2+=∫t12tdt +=2∫t 1t 1++dt-2∫t 1dt +=2t-2ln|1+t|+C=2x -2ln|1+x |+C.(4)∫e sinx sin2xdx=2∫e sinx sinxcosxdx=2∫sinxde sinx =2e sinx sinx-2∫e sinx dsinx=2e sinx sinx-2e sinx +C.(5)∫x e dx=∫e t dt 2=2∫tde t =2te t -2∫e t dt=2e t (t-1)+C=2x e (x -1)+C.(6)方法一：令t-x=1x 2-，则x=2t 1t 2+，dx=222t1t -dt ；∫1x x dx 2-=∫⎪⎪⎭⎫ ⎝⎛++-2t 1t -t 2t 1t 2t 1t 2222dt=2∫1t dt 2+=2arctant+C=2arctan(1x 2-+x)+C. 方法二：∫1x x dx 2-=-∫2x 11x 1d-=arccos x1+C.(7)方法一：∫x tan 1x tan 1+-dx=∫t 1t 1+-darctant=∫)t t)(11(t 12++-dt=∫t 1dt+-∫2t1t +dt =ln|1+t|-21∫22t 1dt +=ln|1+t|-21ln|1+t 2|+C= ln 2t 1|t 1|+++C= ln ttan 1|tant 1|2+++C. 方法二：∫x tan 1x tan 1+-dx=∫x sin x cos x sin x cos +-dx=∫xsin x cos x)sin x (cos d ++=ln|cosx+sinx|+C.(8)∫32)2-x (x-x dx=∫2-x dx +3∫22)-(x dx +2∫32)-(x dx =ln|x-2|-2-x 3-22)-(x 1+C.(9)∫xcos dx 4=∫sec 2xdtanx=∫(tan 2x+1)dtanx=31tan 3x+tanx+C. (10)∫sin 4xdx=41∫(1-cos2x)2dx=41(∫dx -∫cos2xd2x+∫cos 22xdx)=41x -41sin2x+81∫(cos4x+1)dx=41x -41sin2x+321∫cos4xd4x +81∫dx =41x -41sin2x+321sin4x +81x+C=83x -41sin2x+321sin4x +C. (11)由4x 3x 5-x 23+-=)1x (2)-x (5-x 2+≡2-x A +22)-x (B +1x C+得：x-5≡A(x-2)(x+1)+B(x+1)+C(x-2)2.当x=2时，B=-1；当x=-1时，C=-32；由A+C=0，得A=32.∴∫4x 3x 5-x 23+-dx=32∫2-x dx -∫22)-x (dx -32∫1x dx +=32ln 1x 2-x ++2-x 1+C. (12)令t=1+x ，则x=t 2-2t+1，dx=2t-2. ∫arctan(1+x )dx=xarctan(1+x )-∫xdarctan(1+x )=xarctan(1+x )-∫]1)x 1[(x 2x2++dx=xarctan(1+x )-∫)1t )(1-(t 21)-1)(t 2t -2(t 22++dt =xarctan(1+x )-∫1t 1)2t -(t 22++dt=xarctan(1+x )-∫dt+∫1t dt 22+=xarctan(1+x )-t+ln(t 2+1)+C=xarctan(1+x )-1-x +ln(x+2x +2)+C =xarctan(1+x )-x +ln(x+2x +2)+C 1.(13)∫2x x 47+dx=∫2x x 2x 437++dx-∫2x x 243+dx=∫x 3dx-21∫2x dx 44+=41x 4-21ln(x 4+2)+C.(14)∫x tan tanx 1tanx 2++dx=∫2tt 1t++darctant=∫)t (1)t t (1t 22+++dt =∫2t 1dt +-∫2t t 1dt ++=arctant-32∫131t 32t32d2+⎪⎪⎭⎫⎝⎛+=x-32arctan ⎪⎪⎭⎫ ⎝⎛+31t 32+C =x-32arctan312tanx ++C.(15)方法一：∫1002x)-(1x dx=991∫x 2d 99x )-(11=991[992x)-(1x -∫992x)-(1dx ]=991[992x)-(1x -2∫99x )-(1x dx]=992x)-99(1x -98992⨯∫xd 98x )-(11 =992x)-99(1x -98x )-98(1992x ⨯+98992⨯∫98x )-(11dx =992x)-99(1x -98x )-98(1992x ⨯+9798992⨯⨯∫d 97x )-(11=992x)-99(1x -98x )-98(1992x ⨯+97x )-97(198992⨯⨯+C =992x)-99(1x -98x )-(19499x ⨯+97x )-97(194991⨯⨯+C. 方法二：∫1002x)-(1x dx=∫1002x)-(1x)-(1dx-2∫100x )-(1x )-(1dx+∫100x )-(1dx=∫98x )-(1dx dx-2∫99x )-(1dx +∫100x )-(1dx =97x )-97(11-98x )-49(11+99x )-99(11+C. (16)令arcsinx=t ，则x=sint ，dx=costdt.∫2x arcsinx dx=∫tsin tcost 2dt=-∫td sint 1=-sint t +∫sint 1dt=-sint t +∫2t 2tan2t sec 2dt =-sint t +∫2t tan 2t dtan dt=-sint t +ln|tan 2t |+C =-sint t +ln sint cost -1+C =-xarcsinx+ln x x -1-12+C.(17)∫xln ⎪⎭⎫⎝⎛+x -1x 1dx=21∫ln ⎪⎭⎫ ⎝⎛+x -1x 1dx 2=21x 2ln ⎪⎭⎫ ⎝⎛+x -1x 1-21∫x 2dln ⎪⎭⎫ ⎝⎛+x -1x 1 =21x 2ln ⎪⎭⎫ ⎝⎛+x -1x 1-∫22x -1x dx=21x 2ln ⎪⎭⎫ ⎝⎛+x -1x 1+∫dx-∫2x -1dx=21x 2ln ⎪⎭⎫⎝⎛+x -1x 1+∫dx-21ln ⎪⎭⎫ ⎝⎛+x -1x 1+C=21(x 2-1)ln ⎪⎭⎫⎝⎛+x -1x 1+∫dx+C. (18)∫xsinx cos dx 7=2∫31)x sin2x(cos2d2x +=2∫322221t 1t -1·t 1t 2t 12⎪⎪⎭⎫ ⎝⎛++++dt=∫tt 12+dt=∫t1dt+∫3t dx=2t +525t +C=2tanx +52x tan 5+C.(19)∫e x 22x 1x 1⎪⎭⎫ ⎝⎛+-dx=∫e x 222)x (1x2x 1+-+dx=∫2x x 1e +dx-2∫22x )x (1e +dx=∫2x x 1de ++∫e x d 2x 11+=2x x 1e +-∫e x d 2x 11++∫e x d 2x 11+=2x x 1e ++C. (20)I n =∫u v ndx=1b 1∫uv n du=1b 2∫v n d u =1b 2v n u -1b 2∫u dv n=1b 2v n u -12b 2nb ∫v n-1u dx=1b 2v nu -12b 2nb ∫uuv 1-n dx. 又∫uuv 1-n dx=∫ux )v b +(a 1-n 11dx=21b b ∫ux)v b +(a b b 1-n 1112dx=21b b ∫ux)v b +b b a (1-n 2121dx=21b b ∫uv )a b b a (x)v b +a (1-n 21211-n 22-+dx=21b b I n +)b ba a (2121-I n-1 =2b 1[b 1I n +(a 1b 2-a 2b 1)I n-1]. ∴I n =1b 2v n u -1b 2n [b 1I n +(a 1b 2-a 2b 1)I n-1]=1b 2v n u -2nI n +1b 2n(a 1b 2-a 2b 1)I n-1. 即I n =)1n 2(b 21+v n u +)1n 2(b 2n1+(a 1b 2-a 2b 1)I n-1.。