2024年1月10日发(作者：六上期末数学试卷2023)

第八章

一：不定积分概念与基本积分公式（教材上册P181）

1. 验证下列(1)、(2)等式并与(3)、(4)两试相比照:

(1)

f\'(x)dxf(x)c; (2)

df(x)f(x)c;

(3)

[f(x)dx]\'f(x); (4)

df(x)d(x)f(x)dx;

解: (1)

c\'0(f(x)c)\'f\'(x)cf\'(x)

f\'(x)dxf(x)c与(3)相比(1)试求不定积分运算,(2)是求导运算,(1) (3)互为逆运算,不定积分相差一个常数但仍为原不定积分,该常数用c表示,称为积分常数.

(2)

df(x)f\'(x)dxdf(x)f\'(x)dxf(x)c不含积分常数.

与(4)相比: (2)是先求导再积分,因此包含了一个积分常数,(4)是先积分再求导,因此右侧2. 求一曲线y=f (x),使得在曲线上的每一点(x,y)处的切线斜率为2x,且通过点(2,5).

解:

dy2xdxydy2xdxx2c2

将(x,y)=(2,5)代入得: 5=2+c

C=1

该曲线为yx1

2x2sgnx是|x|在(,)上的一个原函数. 3. 验证y2解:

x2x>0时,y’=()\'x|x|

2x2x<0时,y\'()\'x|x|

2x2x2sgnx022limx0 x=0时,y\'limlimx0x0xx02x0x2sgnx0x2

y\'2limlim()0|x|

x0x0x02因此y\'y\'y\'0|x|

x2综上得y\'(sgnx)\'|x|,x(,)

2x2ysgnx是|x|在(,)上的一个原函数.

24. 据理说明为什么每一个含有第一类间断点的函数都没有原函数?

解: 设x0是f (x)的第一类间断点,且f (x)在U(x0)上有原函数F (x),则F\'(x)f(x),xU(x0).从而由导数极限定理得

f(x)limF\'(x)F\'(x0)f(x0)

limxx0xx0f(x)F\'(x0)f(x0).可见f(x)x0点连续,推出矛盾.

同理

limxx0

二: 换元积分法与部分积分法(教材上册P188)

1. 应用换元积分法求下列积分

(1)

(3)

(5)

cos(3x4)dx; (2)

13x2113x22xxedx;

2dxn; (4)

(1x)dx;

2x1()dx; (6)

22x3dx;

(7)

83xdx; (8)

dx

375x

(9)

xsin(2x)4dxdx(11)

; (12)

;

1cosx1sinx22xsinxdx; (10)

dx;

(13)

cscxdx; (14)

x4x4dx; (16)

x1x2dx;

(15)

dxxlnx;

x3x82dx;

x4dx; (18) (17)

(1x5)3(19)

dxx(1x); (20)

cotxdx;

5cosxdx; (22) (21)

(23)

dxexexdxsinxcosx;

2x3; (24)

2dx;

x3x8x22dx; (26) (25)

5(x1)dxxa22(a>0);

dx(a0); (28) (27)

2(xa2)3/2(29)

x51x2dx;

1x3xdx (30)t3x4x11dx.

x11解: (1)

cos(3x4)dxt2x2t

cosd3

11sintcsin(3x4)c

33(2)

xe2x21t1tdx()2e\'d()2

221t1t1t1t22()d()etdt

()e222241t12ece2xc

44dxt2x11t11dln|t|cln|2x1|c (3)2x1t222

tn1(1x)n1cc (4)①当n1时,(1x)dxtdtn1n1ntx1n②当n1时,(1x)dxln|1x|c

n(5)(13x2113x2)dxx1d(3x)3

2x31(3x)1()23dx1arcsin3xc

33

arcsin(6)22x3t12t122x3122x3dx2dccc

22ln22ln22ln2t2x3tt83x(7)3t222232283xdxtd()tdttc(83x)2c

3399t375xdx(8)375x21t33323d()tdttc(75x)2/3c

t55111(9)xsinxdxtsintdttsintt2dtsintdt

22tx2

11costccosx2c

22t2x4(10)dxxsin2(2x)411cottccot(2x)c

2t224sintd21dxt2xd(2t)1x2(11)2dtsectdttantctan()c

21cosx1cos2t2cost2dx1sinx(12)

dx(sec2xsecxtanx)dxtanxsecxc

21sinxcosx111(13)cscxdx

dxdxxxxsinx2xαsincostancos2222xdtan2ln|tanx|c

x2tan2(14)x1x2dx11d(1x2)1x2c

21x2

x1dx(15)4x441x21x2d()arctan()c

x222421()2dxtlnx1t1(16)tdedtln|t|cln|lnx|c

xlnxettx41111155dxdxd(1x)(1x5)2c (17)535353(1x)5(1x)5(1x)10x3112dx8dx4(18)8x24x28x411x422|c

dln|8xx41621122(19)dx11x()dxln|x|ln|1x|cln||c

x(1x)x1x1x(20)cotxdxcosxsinxdxln|t|cln|sinx|c

(21)cos5xdx(1sin2x)2dsinx(12sin2xsin4x)dsinx

sin5x23sinxsinxc

53(22)dxcosxdxdtanxsinxcosxsinxcos2xtanxln|tanx|c

dxexdexdxarctanexc (23)xxx2x2ee1(e)1(e)2x3d(x23x8)dx2ln(x23x8)c (24)2x3x8x3x8tx1(t1)22x22t22t3123dxdtdt((25)t3t3tt2t3)dt

(x1)5

ln|t|23223tcln|x1|(x1)2c

t2x12xd()atxa(26)dxx2a21x()21a1t21dtln|tt21|c1

ln|xx()21|c1ln|xa2b2|c

aa

(27)令xatan,2t2asectdt

dxasec2t11xdtcottdtsintcc

223/23322322(xa)asectaaaax(28)sin5dxdsin(cos42cos21)dcos

cos1x2xsinx5312153222

coscoscosc(1x)1xc

535t3t26t2456(29)dx6tdt6tdt6tdt6tdt

22231t1t1t1xx

ttdt6tdt6tdt6dt6

t67642t1t2dt

677651tt2t36tln||c

51t56121616

661xxx2x6x3ln||c

1751x6(30)x11t1tx11dx2tdt

t1x11

2(t2t2)dtt24t4ln|t1|c

t1

x141x4ln|1x1|c

x41x4ln|1x1|c\'

2. 应用分部积分法求下列不定积分

(1)

(3)

(5)

(7)

(9)

arcsinxdx; (2)

lnxdx;

2xcosxdx; (4)

lnxx3dx;

2(lnx)dx; (6)

xarctanxdx;

a2b2dx(a0).

[ln(lnx)1]dx; (8)

(arcsinx)2dx

lnxsec3xdx; (10)

解 (1)arcsinxdxxarcsinxxdarcsinxxarcsinxxdx

1x

xarcsinx(1x)c

(2)lnxdxxlnxxdlnxxlnxx1221dxxlnxxc

x(3)x2cosxdxx2sinx2xsinxdxx2sinx2xdcosx



x2sinx2xcosx2cosxdx

xsinx2xcosx2sinxc

2lnx112dxlnxdx[lnxx2x2d(lnx)]

x322lnx11

22c2(lnx1)c

2x4x4x1222(5)(lnx)dxx(lnx)x2lnxdxx(lnx)2lnxdx(参考（2）结果)

x(4)

x(lnx)2xlnx2xc

21121x22dx (6)xarctanxdxarctanxdxxarctanx2221x212111xarctanxdxdx

22221x1211

xarctanxxarctanxc

22211111(7)[ln(lnx)]dxln(lnx)dxdxxln(lnx)xdxdx

lnxlnxlnxxlnx

xln(lnx)c

(8)

(arcsinx)dxx(arcsin)x2arcsinx(1x)222212dx

x(arxsinx)arcsinx(1x)d(1x2)

212

x(arcsinx)2arcsinxd(1x)

x(arcsinx)2(1x)arcsinx2dx

x(arcsinx)2(1x)arcsinx2xc

(9) 令Isec3xdx

212221222122

Isecxdtanxsecxtanxtanxsecxtanxdx

secxtanx(1cos2x)sec3xdxsecxtanxIsecxdx

11Isecxtanxsecxdx

221

(secxtanxln|secxtanx|)c

2

(10)Iabdx(a0)x(xa)1222222122xdxx(ax)21222122x2a2a2xa22dx

x(xa)Ia221x2a2dxx(xa)Ia2xd()

ax()21a1111222则Ix(xa)a2221x1d()(xx2a2a2ln|a2x2x|)c

a2x()21a3. 求下列不定积分

(1)[f(x)]f(x)\'dx(1); (2)f\'(x)1[f(x)]2dx;

(3)f\'(x)dx; (4)ef(x)f\'(x)dx.

f(x)解: (1)[f(x)]f(x)\'dx[f(x)]df(x)1[f(x)]1c

1(2)f\'(x)1dx1[f(x)]21[f(x)]2df(x)arctan[f(x)]c(arccot[f(x)]c1)

(3)f\'(x)1dxdf(x)ln|f(x)|c

f(x)f(x)f(x)(4)e

f\'(x)dxef(x)df(x)ef(x)c

三. 有理函数和可化为有理函数的不定积分(教材上册P198)

1. 求下列不定积分

x3x2dx; (2)2dx; (1)x1x7x12

(3)dxdx; (4)1x31x4;

dxx2; (6)(x1)(x21)2(2x22x1)2dx; (5)x3x3111x2x1解: (1)

x1x1x1x3113122dx(xx1)dxxxxln|x1|c

x1x132(2)x2x2x3111

x27x12(x3)(x4)(x3)(x4)x4(x3)(x4)x211dxd(x4)x27x12x4x27x12dx

11d(x4)2x4(2x7)2d(2x7)

2ln|x4|ln|x3|c

(3)设1ABxC

x31x1x2x12则1A(xx1)(BxC)(x1)

(AB)x(BCA)xAC,

则比较两端系数,得B,C21321,A

33dx11x2dx

21x33x1xx12x11112x1312x13d(x1)d()d()

2x12x13x13(3(33)21)21331(1x)212x1arctanc

ln26xx133

11d(d(x)2x11xdxx(4)4dx1112x122x2(x)2xxxx)21(22x11x)2

1arctan(2x1x)c

1211122x11x2x21xxx41dx21dx21222ln|x2x21|c

x2(x)2xx2111x211x31dx4dx4dx

则1x42x12x122x2x2x21

arctanln|2|c

241x8xx21(5)设B1xC1B2xC21A2

2222(x1)(x1)x1x1(x1)222则1A(x1)(B1xC1)(x1)(x1)(B2xC2)(x1)

(AB1)x4(C1B1)x3(2AC1B1B2)x2(C1C2B1B2)x(AC1C2)

比较两边系数得到A11111,B,C1,B2,C2

44422dx1111112d(x1)d(x1)dx

(x1)(x21)24x18x214x2111112d(x1)dx

22224(x1)2(x1)1x11dxdx

(x21)22(x21)2x21dx1111221ln|x1|ln(x1)arctanx(x1)

22(x1)(x1)48241x(x21)1c

4

(6)x21x25dxdxdx

2222(2x22x1)24(2x2x1)2(2x2x1)由课本P193页逆推公式得

14dx1d(2x1)2[(2x1)21][(2x1)21]2[1(2x1)2]2

x2dx

22(2x2x1)1152x15arctan(2x1)c

2(2x1)2142x22x125x35arctan(2x1)c

22(2x2x1)2



2.求下列不定积分

(1)dx53cosx; (2)

dx2sin2x;

dxx2(3); (4)dx;

21tanx1xx(5)dxx2x; (6)11xx21xdx.

1t22x,dxdt,于是有

解: (1)令ttan,则cosx1t21t22dx53cosx121dt14t2dt

1t21t2531t2

1111xd2tarctan(2t)carctan(2tan)c

221(2t)222dxsec2xdxd(tanx)(2)

22222sinx2secxtanx3tanx2

663tanx662arctan(tanx)c

3622tanx12d

dxcosx1cosxsinxsinxcosxdxdx

1tanxcosxsinx2cosxsinx1d(cosxsinx)1

[dx](xln|cosxsinx|)c

2cosxsinx2(3)(4)设x155sint,则dxcostdt

222x21xx2dx15dx(sint)2dt

2251(x)2421455755sint(cos2t)]dttcostsin2tc

288216x2

[

72x12x3arcsin1xx2c

845(5)令x111sect,t,dxsecttantdt

22222dxsecdtln|secttant|c

11(x)2()222dxx2x

ln|(2x1)2x2x|c

1x1t24txdx(6)令t,则

1x1t2(1t2)2dt11x(1t2)244t2x21xdx(1t2)2t[(1t2)2]dt(1t2)2dt

d(1t2)12tdt2td2

2t1t2t211t2

(1t2)2

2t1tln||c

t211t1x211x2ln||c

xx


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