4-4高等數(shù)學(xué)同濟(jì)大學(xué)第六版本

習(xí)題習(xí)題44求下列不定積分:1. x3x3dx;解x3x3dxx32727dx(x3)(x23x9)27dxx3x3(x23x9)dx271x3dx1x33x29x27ln|x3|C.322. 2x3x23xdx;解2x3x23xdx1x23xd(x23xln|x23x10|C.3. x5x48x3xdx;解x5x48dx(x2x)dxx2x8dxx3x1x31x2x8dx32x4x1dxx3x3xdx4.33x311x31x2x8ln|x|4ln|x1|3ln|x1|C.2dx;解3x3dx(1x1 x2)dx(x2x11x112x31)dxln|x1|112x2x1d(x2x32x2x1 2x2x112(x1)2(d(x1)223)22ln|x1|x2x13arctan2x1C.35. 5. xdx(x1)(x2)(x3);解xdx(x1)(x2)(x2 x2 x1 x31(413)dx1(ln|x2|3ln|x3|ln|x1|)C.26. 6. x21(x1)2(xdx;解解x21(x1)2(xdx[112x1 2x1111(x]dx1ln|x1ln|x1 C7.

2 2 x11 ln|x2C1 2 x11 dxx(x2解

dx(1

x1)dxln|x| x2)C.18.

x(x21)dx(x21)(x2x)

x 1x2 2;解 dx (11x1(x21)(x2x) x 2x2

1 1 )dx2x1ln|x|1ln|x1|1

x1dx2 2 x21ln|x|1ln|x1|12x dx1 1 dx9.

2 4 x21 2 x211 1 ln|x| ln|x1| ln(x2 arctanxC1 1 2 4 2dx;(x21)(x2x解 dx (

x1

x)dx(x21)(x2xx2x1 x211 2x1 1 1 1

dx ln(x22 x2x1 2 x2x1 21 1 ln|x2xln(x21) 1 1 1 2 2 2 x2x11 1 3 2x1 ln|x2x1| ln(x2 arctan C.2 2 3 310. 10. 1x41dx;解 1x41dx(x212x1)(x22xdx42x122xdx42x1x2x222x12dx4221(2x 2)2x22x1dx4221(2xx22)2x122dx82[d(x2x22x12xd(x2 2x]1(dxx2 2x14x22x1 x2dx)2x182ln|82ln|x2x22x1|2x142arctan( 2x 2arctan( 2xC.411. x22(x2xdx;解x22(x2xdxx1(x2xdx1x2xdx122x1(x2xdx321(x2xdx1x2xdx121x2x321(x2xdx1 dx,x2x1因?yàn)?x2xdx23而1(x2xdx11(2x1)231d(2x1)2arctan(2x1),333dx[(x1)2(23)2]22由遞推公式dx(x2a2)n 2a2(n(x2a2)n11[x(2n3)dx(x2a2],得1(x2xdx1dx[(x1)2(23)2]221x1(2dx)12x122arctan2x1,2(3)22x2x1 x2x1 3x2x1333所以所以x22(x2xdx121x2x1 2x2x11 2x12arctan2x12arctan2x1C3 3 3 3xx1x2x1 342x1C.312. 12. dx3sin2;解dx3sin2x 4cos2x1dx14tan2x3dtanx141tan2x(32dtanx)212 arctan2tanxC.313. 13cosdx;解13cosxdx1dxxd( )2x x x1cos2 cos2 (1sec2 )2222dtanx2x1arctan2tan2C.2tan2x22或或13cosx令utandxx2312u1u221udux1u2(2)2du12arctanu2C12arctantan22C.14. 12sinxdx;解12sinxdxdxxd( )x x x x x22sin cos sin2 (csc2 cot )222222x或x或dx1sinxcos令utan21212u1u2 1u2du1u2 1u21uxduln|u1|Cln|tan 1|C216. dx2sinxcosx5;解dx2sinxcosx5令utanx2121u251u21u2 1u2du12du 1113(u1)2(du 515C 513tan 25xC.35)23xd(cot )cot2 cot 12xx 1d(cot )2 2x22x 1(cot )2(2 23)22 322cot 23xC.或12sin令utandxx212u1u221udu21u2udu1du(u1)2(23)22x 3dx213C 322tan 23C.15. 1sinxcosx;解dx1sinxcosx 21dxxd(tan )x x xcos2 tan ) 1tan2ln|tanx|C.2222或或dx2sinxcosx5令utanx2121u251u21u2 1u2du12du113du(u1)2(35)23x 5115C 513tan 25C.17. 113xdx;解113xdx令3x1u11u2du3(u11 1u3u3u23u3ln|1u|C33(x233x13ln(13x1)C.2218.( x)31dx;x1解(x)31dx[(x) x1x x2xC.2223x12319. x11x11dx;解x1dxx11x1uu12udu2(u2u12)duu12(1u22ln|uC2(x1)4 x14ln( x1C.20. dxx4x;解dxx4x令xu41u2u4u3du4(u111u)du2u24u4ln|1u|C2 x4x4x)C.21. 1x21. 1xx1xdx;解令1xu,則x1u2, dx1x1u24u du,(1u2)21x x1xdxu1u2111