不定积分典型例题

1不定积分典型例题一、直接积分法直接积分法是利用基本积分公式和不定积分性质求不定积分的方法，解题时往往需对被积函数进行简单恒等变形，使之逐项能用基本积分公式.例1、求dxxxx∫−)11(2解原式＝Cxxdxxx++=−∫−41474543474)(例2、求dxeexx∫++113解原式＝Cxeedxeexxxx++−=+−∫2221)1(例3、求dxxx∫22cossin1解原式∫∫∫+=+=dxxdxxdxxxxx222222sin1cos1cossincossinCxx+−=cottan例4、∫dxx2cos2解原式＝Cxxdxx++=+∫2sin2cos1例5、dxxx∫+221解原式∫∫+−=+−+=dxxdxxx)111(111222Cxx+−=arctan注：本题所用“加1减1”方法是求积分时常用的恒等变形技巧.二、第一类换元积分法(凑微分法)2CxGCuGduugdxxxgdxxfux++====∫∫∫=)]([)()()(')]([)()(ϕϕϕϕ还原求出令凑成在上述过程中，关键的一步是从被积函数)(xf中选取适当的部分作为)('xϕ，与dx一起凑成)(xϕ的微分duxd=)(ϕ且∫duug)(易求.例1、求∫dxxxcostan解原式＝∫∫−=xxxddxxxxcoscoscoscoscossinCxxdx+=−=−∫cos2cos)(cos23例2、求∫−dxxxx2arcsin解原式)()(1arcsin211arcsin2xdxxdxxxx∫∫−=⋅−=Cxxdx+==∫2)(arcsin)(arcsinarcsin2注)(21xddxx=例3、求∫−−dxxx2491解原式∫∫−−+−=−)49()49(81)2(3)2(21221222xdxxxdCxxxxxd+−+=−+−=∫222494132arcsin214941)32(1)32(213例4、求∫+⋅+dxxxx2211tan解原式＝Cxxdx++−=++∫|1cos|ln11tan222例5、求dxxxx∫−−12解原式＝∫∫∫−+=−−−+dxxxdxxdxxxxxx1)1()1(22222Cxxxdxx+−+=−−+=∫2323223)1(313)1(1213例6、求∫+dxxtan11解原式＝∫∫+−+=+dxxxxxdxxxx)sincossincos1(21cossincosCxxxxxdxxx+++=⎥⎦⎤⎢⎣⎡+++=∫|)sincos|ln(21)sin(cossincos121例7、求∫−+−dxxxx11ln112解原式＝Cxxxxdxx+−+=−+−+∫11ln41)11(ln11ln212例8、求∫+dxex11解原式＝∫∫∫+−=+−+dxeedxdxeeexxxxx111Cexededxxxx++−=++−=∫∫)1ln()1(114例9、求∫−+dxeexx1解原式＝Ceededxeexxxxx+=+=+∫∫arctan)()(11122例10、求∫+dxxxsin1sin解原式＝∫∫∫−−=+−dxxxdxdxx2cossin1)sin111(dxxxdxxx∫∫+−=22cossincos1Cxxx++−=sectan例11、求∫−xxdxln32解原式)(ln)ln32(21xdx−∫−=Cxxdx+−+−⋅−=−−−=∫−2121)ln32(121131)ln32()31()ln32(Cx+−−=ln3232例12、求∫+dxxbxa2222cossin1解原式＝∫∫+=+)tan()tan(111)(tantan12222xbadxbaabxdxabCxbaab+=)tanarctan(1例13、求∫++dxxx11645解原式＝∫∫∫+++−=+++−dxxxdxxxxdxxxxx232322226224)(1)(1)(11Cxxdxxdxx++=+++=∫∫33232arctan31arctan)(113111例14、求∫+dxxx)1(18解原式＝∫∫∫+−=+−+dxxxdxxdxxxxx8788811)1(1Cxx++−=)1ln(81||ln8例15、求∫+−−dxxxx54232解原式＝dxxxxxxxd∫∫+−++−+−541454)54(23222∫+−−++−=1)2()2(4|54|ln2322xxdxxCxxx+−++−=)2arctan(4|54|ln232注由于分子比分母低一次，故可先将分子凑成分母的导数，把积分化为形如∫++dxcbxax21的积分(将分母配方，再凑微分).例16、已知2ln)1(222−=−xxxf，且xxfln)]([=ϕ，求∫dxx)(ϕ.解因为1111ln)1(222−−+−=−xxxf，故11ln)(−+=xxxf，又因为xxxxfln1)(1)(ln)]([=−+=ϕϕϕ，得xxx=−+1)(1)(ϕϕ，解出11)(−+=xxxϕ，从而6Cxxdxxdxxxdxx+−+=−+=−+=∫∫∫|1|ln2)121(11)(ϕ例17、求∫dxx4cos1解原式Cxxxdxxxd++=+==∫∫322tan31tantan)tan1(tansec例18、求∫++dxxxx2)ln(2ln1解原式＝Cxxxxxxd+=+∫)2lnarctan(21)ln(2)ln(2三、第二类换元法设)(txϕ=单调可导，且0)('≠tϕ，已知CtFdtttf+=∫)()(')]([ϕϕ，则CxFCtFdtttfdxxfxttx++==−===∫∫−)]([)()(')]([)(1)()(1ϕϕϕϕϕ还原令选取代换)(txϕ=的关键是使无理式的积分化为有理式的积分(消去根号)，同时使dtttf∫)(')]([ϕϕ易于计算.例1、求∫−+221)1(xxxdx解令tdtdxtxcos,sin==原式＝∫∫−−=+ttdtttdtt22cos2coscos)1(sincossintdttcos)cos21cos21(221∫++−−=CxxCtt+−−−+−=+−+−=221212ln221cos2cos2ln2217例2、求∫+241xxdx解令tdtdxtx2sec,tan==原式＝tdtttdttttdttttdtsin)sin(sinsinsinsin1sincossectansec24424342∫∫∫∫−−−=−==⋅CxxxxCtt++++−=++−=)1(3)1(sin1sin13123323例3、求dxxx∫−229解令txsec3=，则tdttdxtansec3⋅=原式＝∫∫∫−==⋅⋅dtttdttttdtttt)cos(secsectantansec3sec9tan3221sin|tansec|lnCttt+−+=12222lnCxaxaaxax+−−−+=Cxaxaxx+−−−+=2222ln例4、求∫+dxxx)2(17解令tx1=，则dttdx21−=，8原式∫∫∫++−=+−=−+=)21(21114121)1(21777627tdtdtttdttttCxxCt+++−=++−=||ln21|2|ln141|21|ln14177注设nm,分别为被积函数的分子，分母关于x的最高次数，当1−mn时，可用倒代换求积分.例5、求dxxxx∫−+1122解令tx1=，dttdx21−=原式∫∫−+−=−−+=dtttdttttt222211)1(11111∫∫−−+−−=22212)1(11ttddttCxxxCtt+−−=+−+−=1arcsin11arcsin22例6、求dxxxx∫−432解原式∫∫∫−⋅=−=⋅−===dttttdtttdttttttxdttdx11211212541051411386121211令∫∫−++=⋅−+−=5554510)111(51211112dtttdttttCttt+−++=|1|ln51251210125510Cxxx+−++=1ln51251256125125659例7、求∫+xedx1解令tex=+1，12−=tex，dtttdx122−=原式＝Cttdttdtttt++−=−=−⋅∫∫11ln11212122Ceexx+++−+=1111ln例8、求∫+dxxxxln1ln解令xtln1+=原式∫∫−=+=dtttxdxx1lnln1lnCxxCttdttt++−=+−=−=∫ln1)2(ln32232)1(2123例9、求dxxx∫++−+1111解令tdtdxtxtx2,1,12=−==+因为原式dxxxxxdxxxx∫∫+−+=+−+=12||ln2122而∫∫∫−+=−=+dtttdttdxxx)111(2121222CxxxCttt+++−+++=++−+=1111ln1211ln210原式＝Cxxxxx+++−+−+−+1111ln214||ln2＝Cxxx+++++−11ln414四、分部积分法分部积分公式为∫∫−=vdxuuvdxuv''使用该公式的关键在于',vu的选取，可参见本节答疑解惑4.例1、求∫dxexx3解原式＝xxxxxxdexexexdexexdex∫∫∫+−=−=63323233Cexeexexxxxx+−+−=66323例2、求∫dxxx2cos22解原式∫∫+=+=xdxxxdxxxcos2161)cos1(21232∫∫−+=+=xdxxxxxxdxxsinsin2161sin21612323∫∫−++=++=xdxxxxxxxxdxxxcoscossin2161cossin21612323Cxxxxxx+−++=sincossin216123例3、求∫dxex3解原式Ceteetdetdtetttttttxdttdx++−==∫∫===66333222332令Ceexexxxx++−=333663332例4、求∫dxx)cos(ln11解原式∫+=dxxxx)sin(ln)cos(ln∫−+=dxxxxxx)cos(ln)sin(ln)cos(ln移项，整理得原式Cxxx++=)]sin(ln)[cos(ln2注应用一次分部积分法后，等式右端循环地出现了我们所要求出的积分式，移项即得解，类似地能出现循环现象的例题是求如下不定积分：∫∫xdxexdxexxββααsincos或例5、求∫++dxxx)1ln(2解原式dxxxxxx∫+−++=221)1ln(Cxxxx++−++=221)1ln(例6、求∫dxxx23ln解原式＝∫∫−−=−=)1(ln3ln)1(ln233xxdxxxxdCxxxxxxxxxdxxxx+−−−−=⎥⎦⎤⎢⎣⎡+−−=∫6ln6ln3ln)1(ln2ln3ln2323例7、推导∫+dxaxn)(122的递推公式解令∫+=dxaxInn)(122∫++−+++=dxaxaaxnaxxInnn12222222)(2)(∫++−++=dxaxnanIaxxnnn122222)(122)(12122222)(+−++=nnnInanIaxx⎥⎦⎤⎢⎣⎡−++=+nnnInaxxnaI)12()(212221⎥⎦⎤⎢⎣⎡−++−=−−11222)32()()1(21nnnInaxxanI例8、推导∫=xdxInntan的递推公式.解∫⋅=−xdxxInn22tantan∫−⋅=−dxxxn)1(sectan22∫∫−−−⋅=xdxxdxxnn222tansectan2122tan11)(tantan−−−−−−=−=∫nnnnIxnIxxd注应用分部积分法可以建立与正整数n有关的一些不定积分的递推公式.例9、已知)(xf的一个原函数是2xe−，求∫dxxxf)('解原式Cexxfdxxfxxfxxdfx+−=−==−∫∫2)()()()(例10、求∫+dxxxx)1ln(arctan2解因为∫+dxxx)1ln(2∫++=)1()1ln(2122xdxCxxx+−++=22221)1ln()1(21所以原式＝∫⎥⎦⎤⎢⎣⎡−++22221)1ln()1(21arctanxxxxd13[]∫⎥⎦⎤⎢⎣⎡+−+−−++=2