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)。 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)；事實上，F(xiàn)(x)C'F'(x)f(x) f(x)的任意兩個原函數(shù)僅相差一個常數(shù)。事實上，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，f(x)被積

2、函數(shù)，x積分變量。顯然f(x)dxF(x)CkdxkxCxdx-x1C11lnxC12、基本積分表(共24個基本積分公式)3、不定積分的性質(zhì)f(x)g(x)dxf(x)dxg(x)dxkf(x)dxkf(x)dx(k0)cscxcscxcotxdxcscxcscxcotxdxesc2xdxcscxcotxdxcotxcscxCdxsin2xcos2x.22sinxcosx,22dxsinxcosx22cscxdxsecxdxcotxtanxC1、2、§2、不定積分的換元法、第一類換元法（湊微分法）faxbdx1faxabdaxb,即dx1-daxba1、求不定積分sin5xdxsin

3、5xd5x5xu1sinudu512x7dxdx2xdx1dx例2、求不定積分2x7d(12x)x.1x2dxx2e3xdx-2cos1dxxxcosxdx、x3、1dxxdxn,x2cos-dxxarcsina即xn1dxx3cosxd、xx2x3dxn.1sinxdlnx,exdxdex,sinxdxsecxtanxdxxdx22.axdsecx,2dx1x2da2x2,tanxdxsinxdxcosx1cos(5x)C512xx22dxx161132x8C(20)(23)3x22C1dxx2dxdcosx,cosxdxdsinx,sec2xdxdtanx,darctanx,dx.1x2d

4、arcsinx,dcosxIncosxCInsecxCcosx(16)cotxdxcosx,dxsinxdsinxlnsinxClncosxCsinxsecxsecxtanxdsecxtanxsecxdxdxlnsecxtanxCsecxtanxsecxtanxcscxcscxcotxdcscxcotxcscxdxdxlncscxcotxCcscxcotxcscxcotx1dxdlnxlnlnxCxlnxlnxdxdtanx12lntanx1Ccosx1tanxtanx1x(17)(18)(19)dxxex1eIn1dxx1exxeex1eIn1exCdex2arctanex1dx1x2d-1

5、x2例4、求不定積分dx2xx22adxd(xa)d(xa)2a丄In2a(21)(22)x1x22-dxx21x-2"x3dxdxdx21x21dx2x丄In2x23arctanx4x22x5dx2x26dx5x22xdx2x22x2x5dxx124sin2xdx1.2lnx2cos2x.dx22x-arctan221cos2xd2x21sin2xC41sin5xcos3xdxsin8x2cosxdxcotx,dxInsinxdx11sin2xdxcos8xcos2xC164dInsinxdsinx1sinxsinInsinx1sinx,2dxcosxsinxInsinxdcosx

6、2cosxsec2xdxInsinxtanxInInsinxCCcosxdxcosxsinxdx2sinxescx4desccot二、第二類換元法1、三角代換例1、.a2aatsin2t24x2dx解：令xasint（或acost）-a2x2acost,dxacostdt原式=acostacostdt21cos2tlxadtdtcos2td2t2a.xCarcsin2a.xarcsinaa22x2C例2、dx22ax.xarcsina解：令xasintacostdt原式=-acostdttC.xarcsina例3、dxa2x2xasect,dxasectdt解：令xatant（或acott），

7、貝Ua2原式=asec2tdtasectsectdtInsecttantInx2Inx.x2a2C例4、dxx一x24解：令xatant(或acott)，貝Ux242sect,dx2sectdttantIn原式=asectdtsectdtInsectasect解:令xasect(或acsct)，貝Ux2x2a2atant,dxasecttantdt原式=asecttantdtsectdtInsecttantCatantxIn-a(25)Inxx2例6、9dxx解：令xasect，貝.x23tant.dx3secttantdt原式=S3secttantdt2tantdt2sect13tant3a

8、rccos-x33arccosCxa22xxasint小結(jié)：f(x)中含有.x2a2可考慮用代換xatant2axasect2、無理代換例7、dx解：令3x1t,則xt31,dx3t2dt原式=rJdtt1dt1ttIndxx1Vx解：令6xt,則x233x13ln1t6,dx6t5dtt21-11FTdt6tarctant66xarctan6x解：令.;xxt,貝収1t21dxt22tdt21原式=t22tdtt212t2t2-dt1；Adt2t-In1C2t12x1x1xIn-1xx例10、十dx_解:令-1ext,貝収Int21,dx2tdtt21Jdtt21I：11e14、倒代換dx例

9、11、xx1解:令x-6,dx4t6dt原式t6dt246d4t14t6124ln4t61C24ln1In41In24§3、分部積分法分部積分公式:UVUV,UVUVUVdxUVdxUVdx，故UdVUVVdU（前后相乘）（前后交換）例1、xcosxdxxsinxcosxC例2、xexdxxxxdexexedxxxe例3、Inxdxxlnxxdln或解：令lnxt,xte原式tdettetetdttetxlnxxxdsinxxsinxsinxdx例4、arcsinxdxetCexCxlnxxarcsinxxdarcsinxxarcsinxxlnxx一xdxv'1x21d1x2

10、xarcsinx2x2xarcsinx.1x2或解：令arcsinxt,xsint原式tdsinttsintsintdttsintcostCxarcsinx1x2Cxlnx.1x2xlnx.1x2XLxi£Zdxxlnx打?qū)弜、1x2.1x2Cx1x2dx例5、exsinxdx例7、lnx,1x2dxsinxdexxesinxxxxecosxdxesinxcosxdexxesinxecosxxedcosxxxesinxcosxesinxdx故exsinxdx1xesinxcosxC2例6、x2-dxcosxxdtanxxtanxtanxdxxtanxInsecxC§4、兩種

11、典型積分、有理函數(shù)的積分P()nn1有理函數(shù)R(x)空2壯an1xa1xa。可用待定系數(shù)法化為部分分Q(x)bmXmbm1xm1bxb。式，然后積分x32x5x6<35x6xB1,2B3x3x32x5x6<35x6xB1,2B3x3例1、將AA3AB5xx22x6x3解：丐x化為部分分式，并計算F3dxx5x6ABx3A2B2x3dxdxdx5ln(x2)6ln(x3)C2x5112dxx5x62xx25x65xdxx25x6lnx25x62例2、例3、例4、!lnx22dxx(x1)25x6dnx32x1xdxx(x1)2Adx1x(x1)dx(x2x4xdx122?dx1丄Jdx12xx2x2x2dxx2arctan212xdx12x1xxC、21-1xdx1x212xarctanJxx2arctanx、2x12ICx.2x1ln22x2x.2x2x二、三角函數(shù)有理式的積分對三角函數(shù)有理式積分IRsinx,cosxdx,2arctanu,u1u,sinx廠cosx廠dx1u1udu,u2u2，u1u21u2Jdu,三角函1u數(shù)有理式積分即變成了有理函數(shù)積分。例5、35cosx解：令utan二貝Vx2arctanu，cosx2dxdu1u原式例6、2彳du1u1u351u2dx2sinxcosx5解：令uxr,tan,