第四章第三節(jié)分部積分法

1、習(xí)題4-31.求下列不定積分1 2 2d(1 - x ) = xarcsin x 亠.1 - x CX1(1) arcs in xdx=xarcs in x -= xarcs in x Ji - x31 x22131 X2 1 - 12 x arctanx，dx x arctanx廠d(x )2(2) In(x2 1)dx=xln(x2 1)-2篤dx=xln(x21)-2 1dxL1 +xI 1 + X 丿2=x In( x 1) -2x 2arctan x C ；x(3) arctanxdx 二 xarctanx2 1 +xx1 1 2 1 2 dx 二 xarctanx2d(1 x )二

2、 xarctanxIn(1 x ) C21 + x22工xxcos dx ,2excos二dx 二-二 exdsin 二= -2exsin二-4 ex sinxdx，代入上式，得2(4) exsin?dx = 2 exdcos =2excos 一4 e2xx2_2x_2x所以2xsindx - -2ex22(5) x arctan xdxcos -8exsin x -16 exsinxdx,2 2 2_2x . x .e sin dx2ex I cos 4sin C ;1722=-arctan xdx33Aarctanx-丄3x32dx3 1 x2-Warctanx-l361d(x2)1d(x

3、26 1 x21 x3 arctanxx236ln(x2 1) C;6xx 1xx(6) xcos dx =2 xcos dx = 2 xdsin2xsin 222 222X .- x 1 .-. . x 2 * 2 2 * 2xx 1xx= 2xsin 4 sin dx =2xsin4sin C;22 222si n? dx2222x(7) xta n xdx 二 x(sec x -1)dx = xd ta nx- xdxxta nx- tan xdx 22x 丄二 xtan x ln | cosx | C ;2 2 2 22xdx2 x -1(8) In xdx =xln x -2 In

4、xdx = xln x -2x1 nx 2 1dx = xln x -2xln x 2x C;(9) xln x -1 dx： ln x 1 dx2x2ln(x1)；22t1 2 1 x ln(x -1)21 i1111(x 1)dx x21n(x-1)x2xln(x-1) C；2 |Lx -12422(10).ln2 x 廠dx x=-ln2 xd丄xl n2x 2 ln xdx-n2x2 ln xd-xxxx1221ln2x ln x C (In2 x 2ln x 2) C ；xxxx(11) cos(l nx)dx = xcos(l nx)亠 i sin (I n x)dxsin(ln

5、x)dx = xsin(ln x) - cos(ln x)dx 代入上式，得cos(lnx)dx = xcos(lnx) xsin(In x) - cos(lnx)dxxcos(l nx)dx cos(l nx) sin (I n x) C ;In x1In x(12) rdx = - Inxd-XXX1(13) xnlnxdxIn xdxn 1n +1 1dx - - x C;xxn dlnx-n 1 x&1 n 11n 1x In x -xn12 x (n 1)21In x -C (n - -1)(n = -1);(14)x2edx_ -x2de =x2e2xedx_-x2e-2 xde=

6、x2e舟-2xe2edx-x2e -2xe -2e C - -e(x2 2xe2) C ;(15) x31n2xdx=In2 xdx4 二= x4lnx1 x31nxdx 二x4ln2x】 In xdx44 4248 L42413.42414142.1x In x x In x x dx x In x x In x x C x 2I n xT nxC48848328.4In (I n x)11(16)dx = ln(ln x)dlnx =1 nx In(ln x) - Inx dx=lnxln(lnx)dxx、xln x、x二 lnx ln(ln x) -Inx C 二 lnx lln(ln

7、x) -1 C ；1 1 11(17) xsin x cosxdxxsin 2xdx xdcos2x xcos2x cos2xdx244411xcos2x sin2x C ；48x111(18) x2 cos2 dxx2(1 cosx)dxx2dx1 21 3 12x cosxdx x x dsin x;2 6 2132132x x si nx xsinxdx x x sinx 亠 ixdcosx 6 2 6 21312x x sinx xcosx-sinx C6 22 1 2 1 2(19) (x -1)sin 2xdx(x -1)dcos2x(x -1)cos2x 亠 i xcos2xck

8、12 112 11(x -1)cos2x xdsin 2x (x -1)cos2x xsin2x sin2xdx； 2 22 221 111 f(x2 -1)cos2xxsin2x cos2x Cx2 2422 31cos2x xsin 2x C2(20)令，則 x 二t3, dx =3t2dt,je扳dx=3Jt2ddx =3t2d -6Jtetdx=3t2et -6td +6Jetdx=3t2et-6上+6+。= 3et(t2t 2) 3ex 3x2-23x7 2 C ;(21) (arcsin x)2dx = x(arcsin x)2 -2arcsinxdx = x(arcsin x)2

9、 + arcsinxd( x2)J1-x2 J1-x2二 x(arcs in x)2 2 arcs in xd、1 - x2 二 x(arcsin x)2 2 1 - x2 arcs in x - 22 2=x(arcs inx)2J_x arcs in x-2x C;(22)因?yàn)閑xcos:2d coxd2;x = ex co s:2sixnx2 d同時(shí)exsi n x x si nx2xed esixi 2xe2ex) x所以excos2:dx lecosx xe si os(2x1(其中CC1);2dx = 2 &ln(1 x) -4 -X2jx(1 + x)dx(23) ln(1以=2

10、 In(1 x)dG = 2&ln(1 x) -2 Jxx= 2 x1 n(1 x)4d.x=2、_xln(1 x)41+x=2 . x ln(1 x) -4 1d , x 4 1 _ 2 d、- x = 2 . x ln(1 x) - 4 . x 4arctan . x C ;1+(仮)(24) . Ldx=e* ln(1 ex)dx - - ln(1 ex)de* = e ln(1 ex).x xe e-dx1 ex-e ln(1ex)1廿dxe5(1 ex).x x1 e -e ,x dx1 ex=-el n(1 ex)x1dx1dx = e ln(1 ex) x：d(1 + ex)1e

11、x二-eln(1 ex) x ln(1 ex) C = -eln(1 - ex) -ln(1 ex) ln ex C -eln(1 ex) Tn(e 1) C ;1 + x11(25) xln dx 二 xln( 1 x)dx - xln(1x)dx ln(1 x)dx2ln(1x)dx21x22x12、1x12,1x1211dxx ln(1 x)dxx Inxu入入 Illi 1 入丿一U入一入 III一|入11IP1 x221-x21-x21x1-xdxIn(1 x) 一12,1 xIn1 -xx2冷 x2ln3 1dx 1_x2lnE 1dxdx-x 1 x=】x2l門(mén)山 xln1(x2

12、 -1)ln1 x C；21-x 2 1-x 21-x= t2ln(1t)-(26)令x -t，則 x =t2,dx =dt2 =2tdt，所以In(1、x)dx = In(1 t)dt2 =t2ln(1 t)-=t2 ln(1 t) 1 t -ln(1 t) C = (t2 1)ln(1 t)-牛 t C = (x 1)ln(1. x) 一； . x C1 2 1 13cscxsec xdx cscx tanx cscxcotxtanxdx2二 2 - -. 11cscxdxln | cscx -cotx | C .2 sin x cosx 22cosx 2dxdx(27) J2sin 2x

13、cosx 2sin xcos x11 sin x 12.已知是f (x)的原函數(shù)，求 xf (x)dx. x解： xf (x)dx = xdf (x) = xf (x) - f (x)dx = xsin xsin xC3.已知 f(x)=求 xf (x)dx.解：xf (x)dx=xdf (x)二 xf (x)-f (x)dx = xf (x) - f (x) Cf X、rxf x、ee丄小ex-一 +C =xi 丿x 1 -x2(arcsin x)n - n(n -1) (arcsin x)n dx=x(arcsinx)n n-x2(arcsinx)n4 -n(n - 1)In.6.設(shè)f(x)是單調(diào)連續(xù)函數(shù)，f J(x)