華東師范大學(xué)數(shù)學(xué)分析第8章習(xí)題答案

1、第八章一：不定積分概念與基本積分公式（教材上冊p181）1. 驗(yàn)證下列 (1)、(2) 等式并與 (3) 、(4) 兩試相比照 :(1)f '(x)dxf ( 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 '( f (0x)c) 'f '(x)cf '( x)f '( x)dxf ( x)c與(3) 相比 (1)試求不定積分運(yùn)算 ,(2) 是求導(dǎo)運(yùn)算 ,(1) (3) 互為逆運(yùn)算 ,不定積分相差一個(gè)常數(shù)但仍為原不定積分 ,該常數(shù)用 c 表

2、示 ,稱為積分常數(shù) .(2)df( x)f '( x)dxdf ( x)f '(x)dxf( x)c與(4) 相比 : (2) 是先求導(dǎo)再積分 ,因此包含了一個(gè)積分常數(shù),(4) 是先積分再求導(dǎo),因此右側(cè)不含積分常數(shù) .2. 求一曲線 y=f (x), 使得在曲線上的每一點(diǎn) (x,y) 處的切線斜率為2x,且通過點(diǎn) (2,5).解:dy2 xdxydy2xdxx2c2將(x,y)=(2,5) 代入得 : 5= 2 +c該曲線為c=1yx213. 驗(yàn)證 y2x sgn x 是|x|在 (,) 上的一個(gè)原函數(shù) .2解:x2x>0 時(shí),y= ()' 2x| x |2xx&

3、lt;0 時(shí), y '()' 2x| x |22x sgnx0x22xx=0 時(shí), y 'limlimlim0x0x0x2x0xx02xsgnx02y '2limlim()0| x |x0x0x02因此 y 'y 'y '0| x |綜上得x2y '(sgn x)'| x |,x 2(,)x2ysgn x 是|x|在 (,) 上的一個(gè)原函數(shù) .24. 據(jù)理說明為什么每一個(gè)含有第一類間斷點(diǎn)的函數(shù)都沒有原函數(shù)?解 :設(shè)x0 是 f(x) 的 第 一 類 間 斷 點(diǎn) , 且 f(x) 在u ( x0 )上 有 原 函 數(shù) f(x

4、), 則f '(x)f (x), xu(x0) .從而由導(dǎo)數(shù)極限定理得limxx0f (x)limxx0f '(x)f ' (x0)f ( x0 )同 理 limxx0f (x)f '(x0)f ( x0 ) .可見f ( x) x0 點(diǎn)連續(xù) ,推出矛盾 .二: 換元積分法與部分積分法(教材上冊 p188)1. 應(yīng)用換元積分法求下列積分(1)cos(3xdx4)dx ;(2)2 x2xedx ;n(3)2 x1;(4)(1x) dx ;(5)(11)dx ;(6)22 x3dx ;3x213x2(7)83xdx ;(8)dx3 75 x(9)x sinx2 dx

5、 ;(10)dx;sin 2(2 xx)4(11)dx1cos x;(12)dx;1sin x(13)csc xdx ;(14)xxdx ;1x2dx(15)4 dx ;(16);4xx ln x(17)x45 3 dx ;(18)(1x )x3x82dx ;(19)dxx(1x);(20)cot xdx ;(21)cos5xdx ;(22)sindx;x cos xdx(23)xxee2 x3;(24)2x3 xdx ;8(25)x22( x1)5dxdx;(26)dxx2a 22x5(a>0);(27)22 3/ 2 (a0) ;(28)dx ;(xa )1x(29)xdx13 xt

6、 3x 4(30)tx11 dx .x11解: (1)cos(3x4)dxcosd31 sin tc1 sin(3x4)c332t 2 x2t1t1(2)xe2x dx() 2 e' d () 222t11t1t1()2 et() 2 d ()et dt22 2241 etc1 e2x2c44dxt 2x 11t11(3)dln | t |cln | 2 x1|c2 x1t222nt x 1t n 1(1x)n 1(4)當(dāng) n1 時(shí),(1x) dxt n dtcc當(dāng) n1 時(shí),(1nx) dxln |1n1n1x |cdx(5)(11)dx31d (3x)3x213x21( x ) 2

7、331(3x)2arcsinx1 arcsin3xc33t 2x 3tt122 x 31 22 x 31 2(6)22 x3dx2t dccc 22 ln 22ln 22ln 2t8 3xt 222 323(7)83 xdxtd ()t 2dtt 2c(83 x) 2c3399dxt3 7 5x1t333232/3(8)d()tdttc(75x)c3 75xt551010t x 2111111(9)x sinx2dxt 2sin tdt 2t 2 sin tt 2dtsin tdt22112costccos xc22(10)t 2 xdx411 cottc1cot(2 x)csin2(2 xx

8、)sin2 td t22442(11)dxt 2 xd (2t )122dtsec2tdttan tcxtan()c1cosx1cos 2t2cost2dx(12)1sin2xdx2(sec xsecx tan x)dxtan xsec xc1sinxcos x111(13)cscxdxdxsin xxx dxxxd tan x2sincostancos22222xtan x2ln | tan|c2x1122(14)dxd (1x )1xc1x221x2x114x4 dx41()2x2d()21x2x22arctan() 42cdxx ln xt ln xet t de1t1dttln | t

9、 |cln | ln x |cx4(1x5 ) 3dx15(11x5 )3dx515(11x5 )3d (1x5)1 (110x5 ) 2x3x82dx141x82dx4281x8 2d1x422161ln |x42x42|c1dxx(1x)( 1x11x)dxln | x |ln |1x |cln |x1x|ccot xdxcos x dxsin xln | t |cln | sin x |ccos5 xdx(1sin 2 x) 2 d sin x(12sin 2 xsin4 x)d sin xsin5 x523sin 3 xsin xcdxsin x cos xcos xdx sin x

10、cos2 xxd tan xtan xln | tan x |cexdxe x1e(ex ) 2xdxde1(ex ) 2arctanexcx22 x3 x3dx 8d (x2x23 x3x8)8ln( x23 x8)cx2(x2t x 1(t1)22t 22t3dx1)5t3dtt3dt(1t2323 )dtln | t |t2cln | x1|2x1t32t2t32( x1)2cdxx21t xxaa 2()ax1d () a1t 2dtln | tt21 |c121ln | xa( x)2a1 |c1ln | xa2b2|c(15)(16)(17)c(18)(19)(20)(21)(22

11、)(23)(24)(25)(26)(27) 令xa tan,t222a sectdtd xas e c2223 / 23td3 t1 c o t t d t1xs2i ntcc( xa )a s e c taaa3a2x2x5xsinsin542(28)1x2dxd sin(cos2cos1)d cos cos1213cos5cos3cosc(1x2) 21x2c22535(29)xt 3dx6 t5dt6tt6dt6 t 6dt6tt4dt13 x1t 21t 21t 2tt 6dt6 t 4dt6 t 2dt6 dt6tdt 1t 26 t 76 t532t6tln | 1t |c751t

12、66651111x6x7x6752x 26 x63ln |1 |c1x 6(30)x11 dxx11tx 1 1t t2 (t1 2tdt12tt2 )dtt214t4ln | t1|cx14 1x4ln |1x1|cx4 1x4ln |1x1|c '2. 應(yīng)用分部積分法求下列不定積分(1)arcsin xdx ;(2)ln xdx ;(3)x2 cos xdx ;(4)ln x3 dx ;x(5)(lnx) 2 dx ;(6)xarctan xdx ;(7)ln(ln x)1ln xdx ;(8)(arcsinx)2 dx(9)sec3 xdx ;(10)a2b2 dx(a0) .解

13、 (1)arcsin xdxx arcsin xxd arcsinxx arcsin xxdx1xx arcsinx1(1x2 )2c1(2)ln xdxx lnxxd lnxx lnxxdxx ln xxxc(3)x2 cos xdxx2 sin x2xsinxdxx2 sin x2xd cos xx2 sin x2 x cos x2cos xdxx2 sin x2 xcos x2sin xc(4)ln xdx1ln xdx 21 lnx x 2x 2 d (lnx)x322ln x112 x24x2c4x2(ln x1) c2212(5)(ln x)dxx(ln x)x 2lnxdxx(l

14、n xx)2 lnxdx (參考（ 2）結(jié)果 )x(lnx) 22x ln x2xc111x2(6)xarc tan xdxarc tan xdx2x2 arctan xdx2221x21 x2 arctan x1dx11dx2221x21 x2 arctan x1 x1 arctan xc22211111(7)ln(lnx)dxln(ln x)dxdxx ln(ln x)xdxdxln xx ln(ln x)ln xcln x xln x(8)(arcsin x)2 dxx(arcsin)2x 2arcsin x (1x2 )12 dxx(arx sin x2)arcsinx(1x2 )12

15、d (1x2 )x(arcsin x)22 arcsin xd(11x2 ) 2x(arcsin x) 22(11x2 )2arcsin x2 dx(9) 令 isec3xdxx(arcsin x)22(11x2) 2arcsinx2 xcis e cx dt a nxs exct ax ntxa nxse cx d xsecx tan x(1cos2x)sec3xdxsecx tan xisecxdxi1 secx tan x1secxdx221 (sec x tan x 2ln | secxtan x |)c11x2a2a22(10) iab2 dx(a10)x(x21a2 ) 22xdx

16、x(a 21x 2 )2dxx2a21xx( x2a2) 2ia2x2a2dxx( x2a 2) 2ia2d()x 2a1111x1()1a則 ix( x2a2) 2a2d()(xx2a2a 2 ln |a2x2x |)c22x 2a2()1a3. 求下列不定積分(1)f (x)f ( x) ' dx(1) ;(2)f '( x)dx ;(3)f '(x) dx ;(4)f ( x)1ef ( x)f (x) 2f '( x)dx .解: (1) f (x)f ( x)' dx f (x)df (x)1 1f ( x)1c(2)f '( x)dx

17、1 df ( x)arctan f ( x)c(arccot f (x)c )1 f (x) 21 f ( x)21(3)f '(x)1dxdf( x)ln |f ( x) |cf (x)f (x)(4)ef ( x ) f'( x)dxe f ( x)df(x)ef (x)c三. 有理函數(shù)和可化為有理函數(shù)的不定積分(教材上冊 p198)1. 求下列不定積分x3x2(1)dx ;(2)x1x27 xdx ;12(3)1x3;(4)1x4 ;(5)dxx( x1)(x1)22 ;(6)(2 x222 x1)2 dx ;解: (1)x3xx31x111x2x11x1x3x1dx(

18、x2x11x1)dx13x312x2xln | x1|c(2)x2x2x311x27 x12(x3)(x4)( x3)( x4)1x4( x3)( x4)x21x27x12dxx4d (x4)x217 x112dx1x4d ( x4)2(2 x7) 2d (2 x7)2ln | x4 |ln | x3|c(3) 設(shè)x13a1x1xbxc2x1則1a( x2x1)(bxc )( x1)( ab) x2( bca)xa ,c則比較兩端系數(shù) ,得 b1 ,c32 , a313dxx3131x1x1x22x1dx2 x113112 x13x1d (x1)3(2 x31d ()112 x12xd ()3

19、) 2133(3)2131(16xln2x) 2x11 arctan 32x13cdxdx211d( x1 )x1d (x )x1x2x12(4) 4dx x1dxx221(x1 )222 x1xx(x )212x1 arctan(1x )c12211x21 dxx4112xdx21x21( x2x21 ) 221ln | 2 2x2 xx2x21 |c21xx11x211x31則dxdxdx 1x42x412x4122 x2x2x21arctan41x2ln |8|cx2x211ab1 xc1b2 xc2(5) 設(shè)22222( x1)( x1)x1x1( x1)則1a( x21)2(b1xc

20、1)( x1)(x21)(b2 xc2 )( x1)( ab ) x4( cb) 3x( 2 acbb2)x(ccb)b(xac ) c11111212121211111比較兩邊系數(shù)得到a, b, c1, b2, c244422dx11d( x1)11d( x21) 11dx( x1)(x21)24x18x214x21114( x21)2d (x21)112( x21)2 dx1dxx11dx( x21)22( x21)2x21dx1 ln | x1|1 ln( x21)1 arctan x1 ( x21) 1( x1)(x21)248241 x( x241) 1cx21x25dx(6) 22

21、 dx22 dx22(2 x2 x1)4(2 x2 x1)2(2 x2 x1)由課本 p193 頁逆推公式得1d (2 x1)4 dx212222 2(2 x1)1(2 x1)11(2 x1) x2(2 x22 x1)2 dx1152 x15 arctan(2 x1)c2 (2 x1)214 2 x22x125 x35 arctan(2 x1)c2(2 x22 x1)22. 求下列不定積分dxdx(1)53cosdx(3);(2)x;(4)2sin 2 x ;x2dx ;1tan x1xx2dx11x(5);(6)x2xx21dx .x解: (1) 令 ttan x ,則 cos x 21t 21t2,