复习指南2006版答案详解第三章不定积分

一.求下列不定积分 1x2ln1x

第三章1 1 1x.解.1 1 1

1 1x 4 1x 1 11 1 1 1 1 1解.1x2arctan1xdxarctan1xdarctanxarctan1xdarctan11 1x 1=2arctan1x (直接计算可知darctanxdarctan1x cosxsinx11sinx(1cos 1cos2cosxsinx11sin 1sin 1sin 11sinx2解.

(1cosx)2 1cosxdx1cosxd1cosx cosxcx(x8解.方法一:x1

dt t7dt1ln(1t8) x(

1

t8 t

8=1ln11 x8 71 方法二:x(x8 x8(x8 x( x8

d(1x8 1=x81 ln|x|8ln(11sin

)c= 8 1sinxcosx

x sin sin2cos2 解.

1sinxcos

dx 22

cos2

2cos22

dx

2cos2

2dcos=1dx

1xln|cosx|cos 2二.求下列不定积分 (x(x1)2 d((x1)2 costdt x22x2

令x1tan

cos2tan2tsecxx1x

=sin2

x解.xtanxx1xcos2

cos3

dsin

dsin

tan4tsectsin4tdtsin411x2=

sin2

3sin3

sin3.

3x 3x(2x2解.xtan

sec2 dsin(2x21)1x (2tan2t1)sectdt2sin2tcos2tdt(2x21)1x=arctansintc 4.

x

(a>a2x 解.a2x a2a2

a2sin2tacostdta21cos2tdt1a2t1a2sin2tcacost a2 = arcsin 5.

2 解.xsin (1x2)3dxcos4tdt(1cos2t)2dt12cos2tcos22t =4t4sin2t8(1cos4t)dt8t4sin2t32sin4t=3arcsinx1sin2t(11cos2t) 412sin2 arcsinx8

2sintcost( ) =3arcsinx1x1x2(52x2) x 1解.xt 1 dx t

dt dt令tsinusinucos t2=1cos3uc x x1解.xt

2t 2x2 1tx2 1tt t

1dt

t1x

t1tx2令tsin sinu1cosudu=cosuuc1tx2cos 三.求下列不定积分e3xe4xe2xe3x exe d(exex 解.e4xe2x1dxe2x1e2xdx(exex)21arctan(e

)2x(14x解.令t2x

dx

tln

1

dt

arctant ln2t 1t2 tln ln =ln

(2xarctan2x)四.求下列不定积分x5x1.(x2)100 解.(x2)100dx99xd(x 99(x2)9999x(x x5x=99(x

99

9998(x

5

(x

98

5x

5

54 99(x2)999998(x2)98999897(x2)9799989796(x 543 5432 c9998979695(x 9998979695(xx1x1x1解. 令x1/t

t4tt t4t sec2 令ttanu2

du ln|tanusecu|c x五.求下列不定积分xcos2解.xcos2xdx1x(1cos2x)dx1x21xdsin 1x21xsin2x1sin 1x21xsin2x1cos2x sec3解.sec3xdxsecxdtanxsecxtanxtanxsecxtan=secxtanx(sec2x1)secxdxsecxtanxln|secxtanx|sec3sec3xdx1secxtanx1ln|secxtanx| 3.

(ln (ln 解. dx(lnx)dxx(lnx) x (lnx)33(lnx)2 6ln (lnx)33(lnx)26lnx x x2(lnx)33(lnx)26lnx6 4.cos(ln解.cos(lnx)dxxcos(lnxsin(lnx)dxx[cos(lnxsin(lnxcos(ln=cos(lnx)dxx[cos(lnx)sin(lnx)]2xcos45.

2dx2dx解2dx

4 x

4 dx4

xcos 2 sin3

8

3xcos3

sinx

2 21xcscxdcscx1xdcsc2x1xcsc2x1csc2x =1xcsc2x1cotx 六.求下列不定积分xln(x1x2 (1x2 xln(x1x2 解.

2ln(

1x2

1x=1ln(x1x2 1 1 211令xtan ln(x

1x2)1 1sec2tdt2(1x2 1tan2ln(x1x2) cos

sec 212sin2t12ln(x1x2) d212 12sin214ln(x1x14=2(1x2

=ln(x1x2)

22(1x 2xarctanx1x1x

xarctan

dxarctan arctanx

1xx21x1x1x1 arctanx dx1x1x1x13.

arctane2 arctan

1

解.

dxe2

arctane2

2

arctanex

1e2x1e2xarctanex1

exdx1e2xarctanex

1 2x212

1

e(1 arctan2

2(ex1e2x)dx2 arctane arctanx)xln(1x2)七.fx)x22x

xx

,求fx)dx解.f(x)dx

(x22x1x2ln(1x2)1[x2ln(1x2)]3x x x考虑连续性,c=－1+ c1=1+1x2ln(1x2)1[x2ln(1x2)]3x xf(x)dx x八.fex)asinxbcosx,(a,b为不同时为零的常数),解.令tex，xlnt,ft)asin(lntbcos(lnt,xf(x)[asin(lnx)bcos(lnx [(ab)sin(lnx)(ba)cos(lnx)]2九.求下列不定积分3x23x(2x解.3x23x(2x3)dx3x23xdx23)

x2cln3(3x22x5)2(3x 2解.(3x22x5)2(3x1)dx (3x22x5)2d(3x22x2ln(xln(x1x21x

=1(3x5

2x5)2ln(x1x21解. ln(xln(x1x2111

1x2) [ln(x121

)]2(1x21x2)ln(11x2dln(11dln(11x2ln(11x2 (1x21x2)ln(11x2=ln|ln(11x2)|十.x0时,fx连续,求

xf'(x)(1x)f(x)dxx2解.

xf'(x)(1x)f(x)dx2 2

xf'(x)f(x)dxx2

f(x)dxx=exdf(x)x

f(x)dx=exf(x)+f(x)dx

f(x)dx=

f(x)

十一.f(cosx2)sin2xtan2x,解.令tcosx2cosxt2,f'(t)1cos2x cos2

1(t2)2 (t 所 f(x)(x

(x2)2dx3(x2)x2十二.求下列不定积分1.xarctanx1.xarctan 解.(1x2)2dx2arctanxd1=arctanx12(1x2

(1x2arctan sec2 arctan 令xtant2(1

x2)

2sec4

2(1x2)

2cos=arctanx1t1sin2t=2(1x2 arctan=2(1x2

1arctanx4

4(1x22. x2.1

x1u dxx1解.u

,x1u2

xdxarcsinu du1 令usin t2sintcostdt2 cos4

sincos3

dttdtan2=ttan2ttan2tdtttan2t(sec2t1)dtttan2ttanttx1x=x x1x3.

arcsinx 1111t

arcsin

arcsinx解.

dxx21

dx+令xsin

costdt+1(arcsinx)2tdcott1(arcsinsin2tcos =tcottcottdt1(arcsinx)2tcottln|sint|1(arcsinx)2 = 1x2arcsinxln|x|1(arcsinx)2 arctan x2(1x2)arctan

dx arctanx1 dx x2arctanxdx arctan解.

x2(1x2

x1

1 =arctanxdx1arctan x1arctanx dx1(arctanx)2x(1x2 =x1arctanx1 dx1(arctanx 1x2 x = arctanx2ln1x22(arctanx)十三.求下列不定积分1.x34x2解.x2sin 4x2dx32sin3tcos2tdt32(1cos2t)cos2tdcos=3

tc

1(4x5

)2

4(4x3

)22.

dx(ax2x2a解.xasec

atan dx(a0)asectasecttantaxx2a

tdtatantat=ex(1ex3.

a xex(1ex e21 解. d1 111

—2

dx=arcsinex 4. 4.2a2ax 2ax解.x2a

令u du令u 2asint8asin=8a =8a4

dt2a2(12cos2tcos2=

2t2a

sin2t2a2

2

dt

2t2a

sin2ta4

sin4t=3a2t4a2sintcosta2sintcost(12sin2t)=3a2t3a2sintcost2a2sin3tcostx 2a x 2a 十四.求下列不定积分sin 1cos解. u sinxsin2xcos2x1u2(u21cos u2(u22)du

u2(u2=1 dudu 12=112u2sinxcos

u22

22211cos

u212 12 2sinxdx2 dx d(2cosx)2cosx 2cosx 令tanx

1t

ln|2cosx|2

ln|2cosx 21t1t

3t43=3

arctantln|2cosx|c4arctan1(tanx)ln|2cosx|2sinxcosxsinxcos 解 sinxcosxdx 12sinxcosx1 sinxcos sinxcos (sinxcos)2 =2 sinxcos dx2(sinxcosx)dx2sinxcosxd(x=1(sinxcosx) 2 sin(x (sinxcosx) ln| )| 十五.求下列不定积分 1 解.

dx

3x dx2d(13

2) 1x

2.

e