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(chēng)F(x)為f(x)的原函數(shù)。 連續(xù)函數(shù)一定有原函數(shù)； 若F(x)為f(x)的原函數(shù)，貝UF(x)C也為f(x)的原函數(shù)；事實(shí)上，F(xiàn)(x)C'F'(x)f(x) f(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Fi(x)Fi(x)'Fi'(x)F2(x)f(x)f(x)0，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(chēng)為f(x)的不定積分，記為f(x)dx，積分號(hào)，f(x)被積

2、函數(shù)，x積分變量。顯然f(x)dxF(x)C例1、求下列函數(shù)的不定積分kdxkxCxdxInxC12、基本積分表(共24個(gè)基本積分公式)3、不定積分的性質(zhì) f(x)g(x)dxf(x)dxg(x)dx kf(x)dxkf(x)dx(k0)例2、求下列不定積分dx21(2)11-2xdxxCCx(2)1xdxx12dxx12)1(12)1C2、x5.1xdx5arcsinx3arctanx2xdxdxlne1lncscxcscxcotxdxcsc2xdxcscxcotxdxcotxcscxCdx_2sin2xcoscot2xdxsin2x22sinxcos2cosx,dxxcsc2xdxsec2

3、xdxcotxtanxC2cscx1dx§2、第一類(lèi)換元法（湊微分法）cotxx21arctanx不定積分的換元法1、faxbdx1faxabdaxb,即dx1-daxba1、求不定積分sin5xdxsin5xd5x5x1usinudu=512x7dx2x7d(12x)1cos(5x)5丄1712x71C162x8Cdx2x1xarctanaa(20)dxxarcsina(23)2、1dxdxn,即xn1dxdxn例2、求不定積分x.1x2dxx2x2*1x23、x2ex'dx13-x3Iedx31e3x3C11.2cosdxcosd1sin1Cxxxxxcosx,dx、x2

4、cosxd、x2sinxC1dxXXdlnx,edxde,sinxdxsecxtanxdxdsecx,2dx1x2-=dx2dwGxdcosx,cosxdxdsinx,sec2xdxdtanx,darctanx,1-dxTF7darcsinx,axdx22xd、a22x,例3、求不定積分sinxdcosxtanxdxdxcosxcosxcosx,dsinxcotxdxdxsinxsinxsecxsecxtanxsecxdxdxsecxtanxcscxcscxcotx一cscxdxdxcscxcotxdxdlnxlnlnxCxlnxlnxdxdtanx1cosx1tanxtanx1IncosxC

5、InsecxCIntanx1CxInsinxCIncosxCdsecxtanx,Insecxtanxsecxtanxdcscxcotx,Incscxcotxcscxcotx(16)(17)C(18)C(19)d1ex1exIn1exCdxx1ex1exex1xexlnieCdex2xdxex2arctanexCxe1dx1x2x2d1x2例4、求不定積分x42x2xx42x2x-dx5dx21x21dx2x1In23arctanxC2x26dx52xdx2x22x2x5dxx1212Inx2cos2x.dx212xsin2xdx-arctan221cos2xd2x2sin5xcos3xdxsi

6、n8x2cosxdxsin2xdxcotx,dxInsinxdxdsinx1ccos8x16dInsin1x21Ccos2xC4sin2x41sinxsinInsinx1sinx2dxcosxsinxInsinxdcosxsec2xdx2cosxdxcosxsinxdx2sinx4cscxInsinxtanxInInsinxCcosxcscx一4cotdx111dx1d(xa)d(xa)Vi/22UAxa2axaxa2axaxa丄InxaC(21)(22)2axax2x2x21x3x32dx2dx12dx1x21x1x2、第二類(lèi)換元法1、三角代換例1、.a2x2dx解：令xasint(或aco

7、st)，貝U/22-axacost,dxacostdt原式=acostacostdt原式=acostacostdtdt1cos2td2t222a.xCarcsin2ax.a2x2Cxarcsina1xa22x2Cdx.xarcsina解:令xasint原式=竺沁acostdttC.xarcsina例3、dx.a2x2解：令xatant(或acott)，則asect,dxasectdt原式=竺辿sectdtasectInsecttantInx2a2例4、Inxx2a2CdxXIx2解：令xatant(或acott)，則x242sect,dx2sectdt(24)asectdt丄lx原式=sect

8、dtasectInsecttant_x2a2In-例5、dx_一x2a2解:令xasect(或acsct)，則x2a2atant,dxasecttantdt原式=asecttantdtatantxsectdtInsecttantCIna-x2a2Inxx2a2(25)例6、dxx解：令xastect，則.x293tant.dx3secttantdt3tant原式=3secttantdt32tantdt3sec2t133sect.x293亠/233arccos-Cx293arccos-3xx22axxasint小結(jié)：f(x)中含有22.xa可考慮用代換xatant22xaxasectCtantt

9、C2、無(wú)理代換例7、3dx1Vx1解：令3x1t,則xt31,dx3t2dt原式=3t2dt1tdt3t33x12233x13ln1Vx1dx解：令6xt,則xt6,dx6t5dt6£-7dtt2dt6tarctantC6VxarctanVx例9、1xdx解:令t,則XFt2-,dx1xx2tdt22t21例10、dx22t212二dtt21In1xx21x1ex解:令1ext,貝収Int21,dx2tdtt214、倒代換dx例11、jxx原式2tdtIn1ex1.1ex1則xx6t6dt1In4124分部積分公式:例1、xcosxdx6,dx4t66d4t14t611In24UVU

10、Vdx§3、dtt2丄In244t6丄In246x6x分部積分法UV,UVUVUVUVdxUVdx，故UdVUVVdU（前后相乘）（前后交換）xdsinxxsinxsinxdxxsinxcosxC例2、xexdxxxxdexexxedxxeexC例3、InxdxxlnxxdInxxlnx1.xdxxxlnxxC或解：令I(lǐng)nxt,xte原式tdettetetdttetetCxlnxxCxarcsinxxdarcsinxxarcsinxx1x2dxxarcsinx1 d1x22 Tx2xarcsinx例4、arcsinxdx或解：令arcsinxt,xsint原式tdsinttsint原

11、式tdsinttsintsintdttsintcostCxarcsinx,1x2C例5、exsinxdx例5、exsinxdxxxsinxdeesinxxxxecosxdxesinxcosxdexxesinxecosxxxesinxecosxxxedcosxesinxcosxexsinxdx故exsinxdx1ex2sinxcosxC例6、x2dxcosxxdtanxxtanxtanxdxxtanxInsecxC例7、lnx1x2dx1xx一2xlnx2xx1x2dxxlnx.1x”1x2,xdx,1x2xlnx12x1x2C§4、兩種典型積分一、有理函數(shù)的積分有理函數(shù)R(x)有理函

12、數(shù)R(x)P(x)Q(x)nn1anxan1xa,xa0bmXmbm1Xm1biXbo可用待定系數(shù)法化為部分分式，然后積分或解:例2、例3、例4、x32x5x6<35x6xB1,2B3x3例1、將AA3ABx22xx35x6dx解：二x化為部分分式，并計(jì)算dxdxdxx5x65ln(x2x32)6ln(x3)2x511,dx6x25x2xx25x65x11dxx5x1|n21|n2dxx25x11dx5xMln2x(x1)2x(x1沁h(yuǎn)xx(x1)lnx1(x1)2dx2x4xdx1dx14x-dx12xdxx2arctan12x-dx12x11-2x1xdxarctanV212、2Inx2cx2xarctanJV2x2xx21、2x、三角函數(shù)有理式的積分對(duì)三角函數(shù)有理式積分IRsinx,cosxdx,令utan-,貝Vx22arctanu,sinx-122u1u,2,cosx牙，dxu1udu，u2u11u212u-2uFdu，三角函數(shù)有理式積分即變成了有理函數(shù)積分。dx5cosx解：令uxrttan,貝Vx2arctanu2cosx2u-2,udxdu原式12亠adu1u1u3521u2du丄ln22