第三节分部积分法分布图示★分部积分公式★几点说明★例3★★例1例★例24★例75★例★例8★例6★例9例11★例10例12★★★★例1314★例例15★例1618例★例17★★分部积分的列表法★例20★例21★例19★例22★课堂练习★内容小结4-3习题★内容要点分部积分公式：??(3.1)vdu?udv?uv????(3.2)vdxu?uvuv?dx分部积分法实质上就是求两函数乘积的导数(或微分)的逆运算.一般地,下列类型的被积函数常考虑应用分部积分法(其中m,n都是正整数).nncosmxxsinmxxnxnxcosemxsinmxenmxn)(lnexxxnnnarcsinmxmx等.xxarccosmxxarctan例题选讲?.求不定积分例1(E01)xdxxcos2??x???dv,xdx?du?cosx,令解一??2??222??xxx?????sinx?xdx,cosxxdx?cosxd?cos??222????选择不当，积分更难进行.显然,,u解二令,?xdvcosxdx?dsinxu?,???sin?xdxxxxdsin?xsin?cosxxdx.?cosxx?sin?xC?2x.(E02)求不定积分例2dxxe2xx?deedvdxu?x?,解????xx2xx22x2xxx2xxdex?x?ee?2xe2dxxedx??xde.)2(xe???xeeC?注：若被积函数是幂函数(指数为正整数)与指数函数或正(余)弦函数的乘积,可设幂函数为u,而将其余部分凑微分进入微分号,使得应用分部积分公式后,幂函数的幂次降低一次.?.求不定积分(E03)例3xdxxarctan2??x???dv?d,u?arctanx,xdx解令??2??22222??xxx1xx??????xdarctanxarctanxdx?dx??arctanx?)(arctanxarctanx??d??222222x?1??221x11x???dx1??arctanx??.?C?arctanx)?arctanx?(x??22222x1????3.例4(E04)求不定积分xdxlnx4??x3???dv,xdx?du?lnx,令解??4??4??x1111??3?4443??dxxlnxdx?ln.?x?Cxlnx?xlnx?x?dx??416444??注：若被积函数是幂函数与对数函数或反三角函数的乘积,可设对数函数或反三角函数为u,而将幂函数凑微分进入微分号,使得应用分部积分公式后,对数函数或反三角函数消失.?x.求不定积分5(E05)例xdxsine????xxxxxx解xdxecossinx??dxsindee?exsin?e?d(sinxe)sin??xxxxx)xdxsin?(eecosx?cose??sinxcosxde?e?xxxdx?cosxesin)?e?x(sinxe?x.xcos)?C(sine?sindx?x?2但在两次分部积分,,弦函数的乘积,udv可随意选取)(：注若被积函数是指数函数与正余.以便经过两次分部积分后产生循环式u,中必须选用同类型的,,从而解出所求积分?.求不定积分例6(E06)dxx)sin(ln??解)]xsin(lnx)dx?xsin(lnx)?xd[sin(ln1?dx?xsin(lnx)?xcos(lnx)?x?)]d[cos(lnxx?xcos(lnx)?x?xsin(ln)?dx)??cos(lnx)]sin(lnx?x[sin(lnx)x?.C?cos(lnx)]?x?sin(lnx)dx?[sin(ln)2请读者,灵活应用分部积分法,可以解决许多不定积分的计算问题.下面再举一些例子.悉心体会其解题方法?3.求不定积分例7(E07)xdxsec???23解xx?sectanxsecdtxdxdx?sectanx?secxanx???32xdxsecxdx?x?secxtanx?sec(sec?x?1)dxsecxtanx?sec?3xdxtan?secxx?ln|secx??sectanx|?3把它移到等号左端去，再两端各除以由于上式右端的第三项就是所求的积分,xdxsec1?3，便得2.secxdx?(secxtanx??x|)C?ln|secxtan2xarcsin?.dx求不定积分例8x1?xarcsin??x?xd1?dx?2arcsin解x1??xdarcsin2?1?x2?1?xarcsinx?x?1?dx1?x??2x?arcsinxx1?.?2Cx?2??1?xarcsinxxxarctan?例9求不定积分.dx2x1????xnxarctax????22解??xx?11xdtr?dxacan??????22??x1?x?1???22)?xx1?xdarctanx??(arctan11?22dxx1?1?x?arctanx??2x1?1?2x??x1darctanx?2x1?11???22tdt?secsecdxx?tanttdt.??xC))?C?ln(x?1tan?ln(sect?t22ttan11?x?22原式?.arctanx?ln()?x1?xCx??1??x.求不定积分10(E08)例dxe2,2xt?,x?ttdt,dx?于是解则令????txtttdt?2tde2?etedx?2eetdt?2txttC?e?(1)t?2.?C(2ex?1)?C2e?2te???.求不定积分例11dxx)ln(1?2,?tt?x,x则令解2t???222?2)ttdln(1?t1?x)dx?ln(1?t)dt?ln(1?t)?ln(dt?t)?ln(?t1t?12tdt2??2.C?ln(1?t)??)tln(1?t??t?dt)?(t?1)ln(?t1?t2t?1x.)?x?C??(x?1)ln(1?x23/1xe?.dxI?求例123x131/x则设1先分部积分，后换元.解法,edx,dv?u?3x1331/?2/3x2/3,xv?e?dx,du?x32311/31/3?xx2/3于是dxe?exI??2232dt,3tt,dx?x?于是再设则??31/????tt2tt2t2xtt2.?2(?tet?63e3edx?t?dt?te?tedt3e?6?edt3t?t2?eC)得,代入上式333/13//311x3xx332/23.?C?1)e?3(xCe?x?2)xI?x?e?(?22223,dtdx?3x?tt,则先换元解法2,后分部积分.设te??t2dtI??3tedt?3ttt,dv?edtu?t,...