大學(xué)微積分第五章習(xí)題答案

第五章不定積分（A）yf(x)的導(dǎo)數(shù)等于x2，且x2時(shí)y5，求這個(gè)1．已知函數(shù)函數(shù).解f(x)(x2)dx12x22xCx2，y5代入上式得：C1將f(x)12x22x1xex，并且曲線經(jīng)過(guò)點(diǎn)f(x)xex2．已知曲線上任一點(diǎn)切線的斜率為(0,2)求此曲線方程。yf(x)，則解設(shè)所求曲線為：f(x)(xex)dx12x2exCx0，y2代入上式得：C1將f(x)12x2ex1t的速度為v3t2，且t0時(shí)距離s5，求此sf(t)，則s(t)v3t2，3．已知質(zhì)點(diǎn)在時(shí)刻質(zhì)點(diǎn)的運(yùn)動(dòng)方程．解設(shè)所求運(yùn)動(dòng)方程為：sf(t)(3t2)dt3t22tC2將t0時(shí)，s5代入上式得：C5.1s32t22t54．已知某產(chǎn)品產(chǎn)量的變化率是時(shí)間f(t)atb(a,b是常數(shù))，設(shè)此產(chǎn)品t的函數(shù)P(t)，且t時(shí)刻的產(chǎn)量函數(shù)為P(0)0，求P(t)．P(t)f(t)atb解因?yàn)樗訮(t)(atb)dta2t2btCP(0)0代入上式得：C0，將P(t)a2t2bt。C(x),固定成本（即5.設(shè)生產(chǎn)x單位某產(chǎn)品的總成本C是x的函數(shù)C(0)）為20元，邊際成本函數(shù)為C(x)2x10（元/單位），求總成C(x).本函數(shù)解C(x)(2x10)dxx210xC，1C(0)20代入得：C20，1將C(x)x210x20。6．求下列不定積分：(1)x(x3)dx32513x(x3)dx(x3x2)dxx22x2C解252(2)(t1)3dtt2解(t1)3dtt33t23t131dt(t3)dtttt2t221t23t3lntC12t(3)22xexdx122xexdx(4e)xdx12ln2(4e)xC解(4)e2x1dxex1e2x1ex1dx(ex1)dxexxC解x2(5)x4x21dx解x2x4dx(x21)(1x4)22dx(11x2x21)dxx21x212x1x32arctanxC3(6)xxxdx解xxxdxx8dx8x8C8xxxxC71515153(7)xcos2dx2cos2dx(cosx1)dx1(sinxx)C12x解22(8)cos2xsin2xcos2xdx解cos2xdxcos2xsin2xcos2xsin2xsin2xcos2xdx()dxsin2xcos2xsin2xcos2x)dx（cotxtanx）Csin2xcos2x(11sin2xcos2x7.用換元法求下列不定積分：(1)5(13x)2dx(13x)2dx(13x)2d(1-3x)2(13x)2C1557解321(2)dx(2x3)2解dx1d(2x+3)111(2x3)22(2x3)223C2(2x3)Cx2(3)x32dx1x2解4x32dxx(x21)x2dx(x2)dxx1x21x21x1x2221x2xxdx1x2dxdx11x2ln(1x2)2arctanxC22(x1)edxx22x(4)解(x1)edxed(x22x)1eC122xx22x22xxx22(5)(lnx)21dxx解(lnx)2113dx(lnx)21d(lnx)(lnx)3lnxCx1(6)exdxx21解e111xxdxexd()eCx2x(7)uu25duuu25du1u25d(u25)1(u25)C3解223(8)x2dx3(x35)25解dx1（x5)3d(x5)(x5)C3x35C3x22133333(x35)2x2(9)dxx1解x21dx（x1）d（x1）x1x12（x1）22x1C33(10)2x1dxx2x3解2x1dx1x2x3d（x2x3）=lnx2x3Cx2x3(11)1dxxx2解1dx1dx1dxx1x（）x1xxx2211(x)2dx2arcsinxC(12)dx49x2解dx149x613x13xdarctanC3x1（）226222(13)dx4x26解dx1112x)dx1ln2xCdx(4x2x(2x)(2)42x42x(14)dxx2x6解dx1(11)dx1(lnx3lnx+2)Cdx(x3)(x+2)x2x6x35x+251x3Cln5x+2(15)1dx2x12x1解1dx2x12x1dx2x12x1212x1d（2x1）2x1d（2x1）41（2x1）（2x1）C33226(16)xx3dx1x4解xx3dx1x4dxdx1141xd（1x4）+11x3xdx21x41x421x4471ln(1x4)1arctanx2C42dxx(1lnx)(17)dxx(1lnx)d(1lnx)(1lnx)ln1lnxC解(18)1dx52xx2解1dx16x12（）dx11（）2dx1arcsinx1C66x152xx26sin23xdx(19)sin23xdx1(1cos6x)dx1(x1sin6x)C解226(20)tanxdxx解tanxdx2tanxdx2lncosxCxsinxcosx1sin2x(21)dx8解sinxcosxdxsinxcosxsinxcos2xdxsinxcosxdxx2sinxcosx(cossinx)21sin2x211C(cosxsinx)2d(cosxsinx)cosxsinxxdx(22)xx21解xdxx21d(x21)1（xx21）xdxxdx22xx21=1x31(x21)2C333ecosexdx(23)xecosexdxcosexdexsinexC解xecosxdx(24)sinxsinxecosxdxedsinxeC解sinsinxxcos5xdx(25)解cosxdxcos4xdsinx(1sin2x)2dsinx5(12sin2xsin4x)dsinxsinx2sin3x1sin5xC35sin2xcos5xdx(26)9解sin2xcos5xdxsin2xcos4xdsinxsin2x(1sin2x)2dsinx(sin2x2sin4xsin6x)dsinx1sin3x2sin5x1sin7xC357(27)dxsin4x解sin2xcos2xdx(csc2xcot2xcsc2x)dxcotx1cot3xCdxsinx4sin4x3(28)tan3xdxtan3xdx(sec2x1)tanxdx1tan2xlncosxC解2(29)1dx1cosx解11cosxdx1dxsec2dtanxCxx222x2cos22(30)sin(x3)sin(3x)dx44解sin(x3)sin(3x)dx12cos(2x)cos(4x)dx44211sin(2x)1sin(4x)C222410(31)4arctanxx1x2dx解4arctanxx1x2dx4arctanxdarctanx11d(1x2)21x22(arctanx)21ln(1x2)C2(32)1sin2x2cos2xdx解1sin2x2cos2xdx12112cot2xd2cotx2arctan(2cotx)C2(33)tan(x3)cos2(x3)dx解tan(x3)dxtan(x3)dtan(x3)1tan2(x3)Ccos(x3)22(34)dtetet解dt11（et）2dtetd(et)=arctan(et)Cetet1e2t(35)dxex1解dxe11eeexxxdx1dxxe-1e-1xx1d(ex1)xlnex1xCxe-111ln(x1x2)dx(36)1x2解ln(x1x2)dxln(x1x2)dln(x11x22332ln(x1x)C28.用換元法求下列不定積分：(1)dx2x31t2x3,則x1(t23)，dxtdt解設(shè)21t1dxtdt1dtt12x31tln(1t)C2x3ln(12x3)C(2)x42x3dxt42x3,則x1(t43)，dx2t3dt解設(shè)21x42x3dx（t43)t2t3dt(t83t4)dt21t93t5C14(2x3)94(2x3)5C39595(3)x21xdx12t1x,則x1t2，dx2tdt解設(shè)x21xdx（1t2)2t(2)tdt2(t22t4t6)dt2(1t32t51t7)C3572211(1x)3(1x)5(1x)7C357(4)x1dxxx2tx2,則xt22，dx2tdt解設(shè)x1t231t22dtdx(t22)t2tdt21xx22(t2arctan)C2(x22arctanx2)Ct2222(5)3(1x2)dx2xsint,則dxcostdt解設(shè)sec2tdt1sin2t)2costdt33(1x2)dx（2xtantCC1x2(6)1(1x)dx22xtant,則dxsec2tdt解設(shè)131(1x)dx1sec2tdtcos2tdt11cos2tdtsec4t222112212xC1x2tsin2tCarctanx(7)1dxxx21xsect,則dxsecttantdt解設(shè)1dx1secttantsecttantdttCarccos1Cxx21x(8)x2dx1x2xsint,則dxcostdt解設(shè)sin2t1costcostdtsin2tdt21cos2tdt11x2dx1x2tsin2tC1（arcsinxx1x2）C222(9)1exdx2tt211ext,則xln(t21),dxdt解設(shè)141t21exdx2dt21dt2t1t1212t1t1C21exx2ln(1ex1)C2tln(10)1ex1dx1解設(shè)ex1t,則xln(1t),dxdt1t1111dxdtt1ex1dtlntln1tCt1t1ttClnex1Clne1xCexln1tx9.用分部積分法求下列不定積分：(1)xexdxxexdxxdexxexexdxxexexC解(2)ln(3x2)dx解2x3xln(3x2)dxxln(3x2)xdxxln(3x2)21xxln(3x2)2x23arctanC3(3)xcos2xdx解xcos2xdx1xdsin2x1xsin2x1sin2xdx222151xsin2x1cos2xC24(4)x2sinxdxx2sinxdxx2dcosxx2cosxcosxdx2解x2cosx2xcosxdxx2cosx2xdsinxx2cosx2xsinxsinxdxx2cosx2xsinx2cosx(5)arctanxdx解arctanxdxxarctanx1x2dxxarctanx1ln(1x2)Cx2(6)arcsinxdx解arcsinxdxxarcsinxdxxarcsinx1(1x2)2d(1x2)x121x2xarcsinx1x2C(7)exdxxt,則xt2,dx2tdt解設(shè)exdxet2tdt2tetdt2teedt2(tetet)Ctt2(xexex)C16(8)lnxdxx2解lnxdxlnxd1lnx1x1dlnxxx2x1lnx1dx1lnx1Cxxxx2(9)x3(lnx)2dx解x3(lnx)2dxln2xd(x4)1x4ln2xx42lnx1dx11444x1x4ln2x1x3lnxdx1x4ln2x1lnxd(1x4)424241x4ln2x1124x4lnx1xdx(8ln2x4lnx1)C1x4444x32(10)xsin2xdxxsin2xdxx1(1cos2x)dx1x21xcos2xdx2解241x2xdsin2x1x21（xsin2xsin2xdx）144441x21xsin2xcos2xC1448(11)sec3xdx17解sec3xdxsecxsec2xdxsecxdtanxsecxtanxtan2xsecxdxsecxtanx(sec2x1)secxdxsecxtanxsec3xdxsecxdxsecxtanxlnsecxtanxsec3xdxsec3xdx1(secxtanxlnsecxtanx)C2(12)arctan4x1dx解arctan4x1dxxarctan4x114x124x14xdxxarctan4x111d(4x1)xarctan4x114x1C84x14(13)(arcsinx)2dx(arcsinx)2dxx(arcsinx)2x2arcsinx1解dx1x2x(arcsinx)22arcsinxd1x2x(arcsinx)221x2arcsinx2dxx(arcsinx)221x2arcsinx2xC(14)1arctanxdxx218解1arctanxdxarctanxd(1)1arctanxdx11xx2xx1x21arctanx(x1)dxx1x2x1arctanxlnx1ln(1x2)Cx21arctanxCxln1xx2lnlnxdxx(15)lnlnxdxlnlnxdlnxlnxlnlnxlnxdlnlnx解x1lnxlnlnxdxlnxlnlnxlnxCx(1x2)lnxdx(16)(1x2)lnxdxlnxd(x1x3)解3(x1x3)lnx(x1x3)1dxx33(x1x3)lnx(11x2)dx33(x1x3)lnxxx3C139ln(x1x2)dx(17)19ln(x1x2)dxxln(x1x2)x1x2xln(x1x2)1x2C解dx(3x4)e3xdx(18)解(3x4)e3xdx1(3x4)de3x1(3x4)e3x1e3dx3x3331(3x4)e3x1e3xC(x1)e3xC33x(19)dxcos2xxdxxsec2xdxxdtanxxtanxtanxdx解cos2xxtanxlncosxC(20)xsinxdxxt,則xt2,dx2tdt解設(shè)2xsinxdxtsint2tdt2tsintdt2tdcost22（t2costcost2tdt）2（t2cost2tdsint）2t2cost4（tsintsintdt）2xcosx4xsinx4cosxC20(21)xarctanxdxxarctanxdxarctanxd(1x2)解21x2arctanx1x21xdx11222x1x2arctanx1xx41xdx2xt,則xt2,dx2tdt設(shè)xxdx2t4dt2t411dt2(t2111x1t1t1t2)dt222(1t3tarctant)C2(1x3xarctanx)C33xarctanxdx1x2arctanx1xx1x1arctanxC2622(22)ex(cosxsinx)dxex(cosxsinx)dx(cosxsinx)dex解(cosxsinx)exe（x-sinxcosx）dx(cosxsinx)ex（sinxcosx）dex(cosxsinx)ex(sinxcosx)exex(cosxsinx)dx21ex(cosxsinx)dxexcosxC(23)１１exdxx31t,則x,dx1dt1解設(shè)xtt2１１exdxtetdttdettetetdtx3111teeCexeCttxx(24)xarccosxdx1x2xarccosxdxarccosxd1x2解1x21x2arccosx1x21dx1x2arccosxx1x2(25)xsec2xdx解xsec2xdxxdtanxxtanxtanxdxxtanx+lncosxC(26)xsinxcosxdxxsinxcosxdx1xsin2xdx1xdcos2x4解21xcos2xcos2xdxxcos2x1sin2xC144222x(12sin2x)1sin2xC1xsin2x1x1sin2x142248(27)sinlnxdx解sinlnxdxxsinlnxxcoslnx1dxxsinlnxcoslnxdxxxsinlnx(xcoslnxxsinlnx1dx)xsinlnxxcoslnxsinlnxdxxxsinlnxdx(sinlnxcoslnx)C2(28)xln(x1x2)dx1x2解ln(x1x2)dxln(x1x2)d1x2x1x21x2ln(x1x2)1x2dln(x1x2)1x2ln(x1x2)dx1x2ln(x1x2)xC10.求下列有理函數(shù)的不定積分：(1)dxx(1x)223解dx111x(1x)21dxx(1x)2x1xlnxln1x1xCln1x1x1C(2)61x3dx解131x261x3dx6133)dx(1x)(1xx2)dx6(1x1xx2(242x)dx2ln1x(2x1)3dx1x1xxxx1222ln1xln1xx2313dx(x1)2242ln1xln（xx1）23arctan2x1C23(3)3x31dxx21解3x31dx3x33x13xdx(3x1x23x1)dxx21x2121)dx3x2lnx12lnx1C(3xx1x1224(4)xdxx416解dx(x24)(x24)dx1(xxxx8x24x24)dxx41611x24x24(lnx241lnx24)C1lnC822162x3x22x3(5)dx解2x3dx(2x2)1dxd(x22x3)dxx22x31x22x3x22x3x22x3lnx22x31(11)dxlnx22x31lnx1C4x34x1x3(6)2x1x22x1dx解2x1dxx22x2x21dx1x22x1dxx22x1lnx22x11dx(x12)(x12)111lnx22x1（）dx22x12x12252lnx12Cx12lnx22x14x2(7)dx1x6解dxxdx11(x)31(x)321x22d(x3)1x6321arctanx3C3dxx2(12x)(8)dxx2(12x)21(x12x4)dx解x22lnx12ln12xCxxxx2x1dx(9)3解dxdx1(11x)dxxxxxx1(x1)(x1)2x1x132221lnx11x121dx1dxx21x2221lnx11arctanx1ln(x21)C224(10)1x(1x10)2dx26解1dxx9dx（設(shè)x10(1x10)2x(1x)102x10t,則dt10x9dx）1x(1x)dx1110t(1t)2dt110[1111t(1t)2]dtt1021(lntln1t1t)C1(ln)C1x1x101011x101010a,b為常數(shù),a0）：f(axb)dx11.求下列不定積分（其中（1）解f(axb)dx1f(axb)d(axb)1f(axb)Caa（2）xf(x)dx解xf(x)dxxdf(x)xf(x)f(x)dxxf(x)f(x)C12.求下列不定積分的遞推公式：（1）I(lnx)ndxn解I(lnx)ndxx(lnx)nxnlnn1x1dxxnx(lnx)nnlnxdxx(lnx)nnIn1n1Ix(lnx)nnI(n1,2,)nn127(2)I(arcsinx)ndxn解1I(arcsinx)ndxx(arcsinx)nxn(arcsinx)n1dx1x2nx(arcsinx)nn(arcsinx)d1x2n1x(arcsinx)nn1x2(arcsinx)n(n1)(arcsinx)dxn1n2x(arcsinx)nn1x2(arcsinx)n(n1)In2n1I(arcsinx)ndxx(arcsinx)nn1x2(arcsinx)n1n（n1）I(n2,nn2（B）.1求下列不定積分：（1）dxxlnx(ln2x1)解dlnxlnx(ln2x1)dtt(t21)dxxlnx(ln2x1)(lnx1tdtlnt1ln(t21)Ct212tlnlnx1ln(ln2x1)C228(2)1sinxexdx1cosx解11cossinxxexdxdxsinxexexdx1cosx1cosxxx2sincosxdx2exdxex2x2cos22cos222exdtanxtanexdxx22extanxC;2(3)xearctanxdx(1x2)3解設(shè)tarctanx，則xtant，dxs=ectdt2earctanxdxtantetsec2tdtsintetdtxsec3t(1x2)3sintetdtsintdetetsintetcostdt而etsintcostdetetsintetcostsintetdtsintetdt1et(sintcost)C2earctanxdxx1earctanxC21x2x(1x2)329(4)11xdxx1x1xt,則x1x1t2,dx4t(1)t22解設(shè)dt1t211xdx4t411121t21t2dt4()dtx1x(1t2)(1t2)2(12ln11ttarctant)Cln1x1x2arctan1xC1x1x1x（5）11tanxdx解11tanxdxcosxdxcosxsinxsinxdxcosxsinxcosxsinxxsinxdxxcosxsinxdxcosxdxcosxsinxcosxsinxcosxsinxxd(cosxsinx)cosxsinxcosxdxcosxsinxxlncosxsinx11tanxdx11tanxdx=1x1lncosxsinxC22（6）xcos4x2dxsin3x30解xxcos42dxsin3xx18xcos2dx1xx1dsinxd81x4x2xsin3sin3sin222218x2xcsc2xcsc2dx21xcsc2()cotxCx18242；（7）dx(1ex)2ext,則xlnt,dx1dtt解設(shè)1(1t)21dt[1111t(1t)2dx(1ex)2]dttt11lntln1t1tCxln(1ex)C1ex（8）sinlnxdxx2解sinlnxdxsinlnxd(-1)-1sinlnxcoslnx1dx1xxxxx2-1sinlnxcoslnxd1xx31-1sinlnx（1coslnx1sinlnxdx）xxx2sinlnxdx-1（sinlnxcoslnx）+Cx22x(9)11x2arctan1x1xdx解1arctan1xdxarctan11xxdarctan1x1x1x21x121x21xarctanCarcsinxdx(10)x(1x)解arcsinxdx2arcsinxdarcsinx4(arcsinx)2C3x(1x)3（11）arctanexdxe2x解arctanexdxarctanexd(-1e)2x2e2x1e1dexarctane22xxe2x)e2x(121e2xarctanex1（1-11e2x）dex22e2x1e2xarctanex（--arctanex）+C1122ex32xtanxsec4xdx（12）解xtanxsec4xdxx1dsec4x1(xsec4xsec4xdx)441xsec4x1sin