不定積分公式

1、Ch4、不定積分1、不定積分的概念與性質(zhì)1、原函數(shù)與不定積分定義1：若F(x) f(x),則稱F(x)為f(x)的原函數(shù)。1連續(xù)函數(shù)一定有原函數(shù)；2若F(x)為f(x)的原函數(shù)，則F(x)C也為f(x)的原函數(shù)；事 實(shí)上，F(xiàn) (x) C F (x) f (x)3f(x)的任意兩個(gè)原函數(shù)僅相差一個(gè)常數(shù)。事實(shí)上，由F1Fi FX)fxf J)得Fi(x) F2(x) C故F(x) C表示了f(x)的所有原函數(shù)，其中F(x)為f(x)的一個(gè)原函數(shù)。定義2：f(x)的所有原函數(shù)稱為f(x)的不定積分，記為f ( x)dx ,積分號(hào)，f(x)被積函數(shù)，X積分變量。顯然f xdxF x C() ()kdx

2、 kxCx1C1xdx1In x C1cot2xdx esc2x 1 dx cot x x C2基本積分表(共24個(gè)基本積分公式)cscx esc x cot x dx esc2 3xdxesex cot xdxcot x cscx C-Axsin2xcos2xsin 2 xdx sin2xcos2xesc2xdx sec2xdx cot x tan x C、第一類換元法1、ax b dx例1、求不定積分sin 5xdx1 2x7dxd a2x2dx2、f xnxn 1dx例2、求不定積分x2dxx3dxx2cos x dxa cos、dx2、不定積分的換元法（湊微分法）ax b d axxd

3、 x 5x u 151 2xd (1 2x)cux-a-1 x a*2ad x.bxarcs in 1 x/a2sinudud ax2x1631dx d in x,secx tan xdxx-dxPa 2 x2tan xdx1arc tan1 2x8C(20)(23)f xndx11，即xndxdx11x2 V2dx2123ecos x dsinx2dxcos Jxd你sindx21 xexdx dex, sin xdx1dxd secx, 1 X?d Ja2x2d cosx, cosxdx d sinx,sec2xdxd tan x,arc tan x,、1 x 2 dx d arcsin

4、x,sin x dxd cosIn cos x C In secx C(16)cos xcos xex-dx1 edx111dx1x 2 a22aXax a2aUnX aC2aXax2x 2X2 1x 3f2dx2dx11 xz1XX4d x2-U3 -d2x211cl ( x a) d (x a)x ax a(21)(22)x 31 X 2dxY19In 1 x23 arc tan x Cx22（3）X 412x 2 6-dxx22x 52x22x 51In x22x 52sin2xdxcos2-idxcot xdxcos x dxsin xQ A P Y e A P Yd sin xsin

5、 xIn sin x C( Q A P YIn cos x C(17)(18)tan xdxtan xIn c a c vtan xcsecxcix111 o C V Asecxtan xsecxtan xcot Xesc xdxoe xxcnf x-dxd CS.CXIn esc xcot Xc(19)CSC Xcot XCSC Xcot Xd In xCIn In x1- dxx In xdxd tan xIn tan xcos2X 1tan xtan xdx1 ex1 exIn 1exCdex1 ex1入dxarctane C例4、求不定積分In xexIn 11d x22x5dxdx

6、32x22x5x 1243Xarc tantC221-1 4-cos2xd 2 x丄X_ sin 2xC24sin 5x cos3xdx _1 sin 8x sin 2x dx2_cos8x16cot x dxpneYHYd sin xd In sin xIn sin xsin In sin xsin x In sinxIn sin xdxJdxsec2xdxfl cos；x tan x221 sin xCOS XCOS Xdxdx1CSC X cicosx sin xV2 sin x4241_ cos2x C4In In sin xC_ ccosx二、第二類換元法1、三角代換解：令x a s

7、in t(或a2X21 In esc x石cost),cot Xacost, dxa costdt原式=a cost a costdta 21 COS2tdtdtcos 2td 2t21a2己sin 2t4Xarcsina2 arcsinX2a24例2、dxa2x21 x/a2解：令x asin t原式 二acostdtdt t Ca cost例3、dx-a2x2解：令x a tan t(或a cot t)則Xarcsinxarcs ina2tdt sec(24)原式=asec2tdta sectsectdt In sectIn xx2Va2C例Adxx%x24tan t C In%-x.2_

8、a _x Caa解:令x a tan t(或a cot t)，貝j* ?4原式=nsec? rdf sectdt In secta sect例5、dxJv2n2xa解：令Xa sect(或a csct),貝!JVx2a2a tant，dx a sect tan tdt例6、X*解：令x a sect ,則Px?9 3 tan t,dx 3sect tantdt原式=3 tant3sect tan tdt3tan2tdt3 sec211 3 tan t t C3sectJ X?%A可考慮用代換X Vx2a2X3arccosCx2J 933 arccostdx2tdt2 sec ,2 sectan

9、 tCIn -XJLT_ar_xC原式=asect tan tcltsectdt In sect a tan tIn x V x2a2Ctan t C InxQ 22xacaa(25)小結(jié)：f(x)中含有a sin t atan ta sectaa(24)2、無(wú)理代換1例7、dx原式二3t2dt11122a1dx1,1 1dt例8、dxLX 1解:令x t6，則dx原式=t2dt33 In 1t5dt66t5dt362 dt 61 t16$x arc tanC例9、尸dxX X解：令X t2原式二t21 t2tdt2 I2-J dt 3 U- t In1 t1177t212dt6 t arct

10、ant1 Idt1 In t 1 cil x刑Y Xdx例10、,_Vl ex解：令Vl ext,則x In t21 , dx-34dtt214、原式_1 dt 2 4U2t t21lnl ex、l ex倒代換1例11、dxx x 641 1dt解:令則,dxXt , X X611 4仁t2161原式t6dtLd 4t61ln4t6iCIn x6c14t6244t612424 x644 In xLIn x64 C424 3、分部積分法分部積分公式：UVU VUV, UVUVU V仮9 1、xcos xdx例2、xexdx或解：令I(lǐng)n xU VdxUV dxUdVUVVdU（前后相乘） （前后交

11、換）xd sin x xsin x sin xdx xsin xcos x Cxdexxexexdx xexex例3、In xdxx In x xd In xx In xx In x x C原式tdeeldttelelCx In x x例4、arcsinxdxx arcsin xxd arcs in xxarcsin xx=dxx arcsin x丄x arcsin x1 x2x2或解：令arcsinsin t原式td sin tt sin tsin tdt t sint costx arcsin x 1 x2Cexcosxdx exsin x cosxdexexsin x excosx ex

12、d cosx exsin x cosx exsin xdx故exsin xdx Cxsin x cosx C2X例6、 -xCOS2X1 x24、兩種典型積分、有理函數(shù)的積分式，有理函數(shù)R(x)然后積分。a -Ya一iY111-ai x如可用待定系數(shù)法化為部分分bi x boQ(x)bmxmbmi xm 1例1、將x 32化為部分分式，并計(jì)算x 3,2dxx 5x 6x 5x 6X 3x 3ABA B x 3A 2B解：X25x 6x 2 x3x2 x3x 2 x 3AB 1A53A2B3B6故-x3dx5dx6-dx51n( x 2)6 ln( x 3) Cx25x 6x 2x 3或解:I丄

13、2x 5-LLdx丄 4-6 4-4- - dx-2 x25x 62x25x62 x25x 612_ In x25x 6-U-1-4 dx22x3 x2仮！5、exsin xdxsin xdexexsin xxd tan x x tan xtan xdx x tan x In secx例7、In xdxxln x J 1 xx In x 1 x21 x21例2、例3、例4、in x22dxX(X1)2X21-dxX41X4dxi5x 6 LL ln_x_3 c2x(x 1 )21X(x l)211 P H一dx2XX2X2x2rdxIndxx( X 1)(XX 1-2 x2二、三角函數(shù)有理式的積分對(duì)三角函數(shù)有理式積分I2u,cos X1 U21 u2,dx1 U2sin x數(shù)有理式積分即變成了有理函數(shù)積分。例5、35 cos xdx解：令Utan?,則x1Jl_dx1VXarctan_R sin x, cos x dx、2u1 u2du，故I2 arc tan u，cos x1-u2, dx1 u21-dx1J ln21X _Xtan_ ,則21 u2arc tan u、1 u2,1 u21 u2du3 du1 u23、不定積分的性質(zhì)1f (x) g( x) dx f (x)dx g( X)dx2kf (x)dx k f(x)dx (k 0)原式例6、-解：令u原式1 2-i一