第四章 不定積分習題詳細解答20230919

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習題4-1

1.已知f(x)之一的原函數(shù)為sin3x，求?f?(x)dx.解：?f?(x)dx?f(x)?C?(sin3x)??C?3cos3x?C.2．設[lnf(x)]??sec2x，求f(x).

解:由于[lnf(x)]??sec2x，故lnf(x)?tanx?Cx1(C1為任意常數(shù))，f(x)?Cetan3.若e?x是f(x)的原函數(shù)，求?x2f(lnx)dx.解：f(x)=(e?x)???e?x,f(lnx)??e?lnx??1x2x,?x2f(lnx)dx??2?C

4.若f(x)是e?x的原函數(shù)，求?f(lnx)xdx.解：由于f?(x)?e?x，所以f(x)??e?x?C10,f(lnx)??e?lnx??x?C0,

f(lnx)x??1C0f(lnx)1x2?x,?xdx?x?C0ln|x|?C.5.求以下不定積分：35（1）解：?(1x?5xx)dx??(x?12?5x2)dx?2x?2x2?C

2）解：?(2x?3x)2dx?4x2?6x9x（2ln2?ln6?2ln3?C

（3）.?1?x?x2x(1?x2)dx??[11?x2?1x]dx?arctanx?ln|x|?C(4)解：?(1x2?1?cot2x)dx??1x2?1dx??(csc2x?1)dx=arcsinx?cotx?x?C

解：?10x23xdx??10x8xdx??80xdx?80x(5)ln80?C

(6)解：?sin2x2dx=??12(1?cosx)dx?112x?2sinx?C

?cos2xcosx?sinxdx??cos2x?sin2(7)xcosx?sinxdx??(cosx?sinx)dx??sinx?cosx?C

(8)解：?cos2xcos2xsin2xdx??cos2x?sin2xcos2xsin2xdx??(11sin2x?cos2x)dx??cotx?tanx?C

(9)解：?secx(secx?tanx)dx??sec2xdx??secxtanxdx?tanx?secx?C??x,(10)解：設f(x)?max{1,x},則f(x)??x??1?1,?1?x?1.

??x,x?1?f(x)在(??,??)上連續(xù)，

1

?12?x?C,x??1?21則必存在原函數(shù)F(x)，F(xiàn)(x)???x?C2,?1?x?1又?F(x)須四處連續(xù)，有

??1x2?.?2C3,x?1xlim11??1?(x?C2)?xlim??1?(?2x2?C1)，即?1?C2??2?C1,lim(1x?1?2x2?C3)?lim(x?1?x?C2)，即12?C3?1?C2,聯(lián)立并令C,可得C11?C2?2＋C,C3?1?C.

???122x?C,x??1?故?max{1,x}dx???x?1?2?C,?1?x?1.

??12?2x?1?C,x?16.解：設所求曲線方程為y?f(x)，其上任一點(x,y)處切線的斜率為

dydx?x3，從而y??x3dx?14x4?C由y(0)?0，得C?0，因此所求曲線方程為y?14x4.7.解：由于

????1sin2x???sinxcosx，???1cos2x???cosxsinx

?2??2?????1??14cos2x???2sin2x?sinxcosx

所以12sin2x、?12cos2x、?14cos2x都是sinxcosx的原函數(shù).

習題4-2

1.填空.

(1)1x=d(?1x+C)(2)12dxxdx=d(lnx+C)

(3)exdx=d(ex+C)(4)sec2xdx=d(tanx+C)(5)sinxdx=d(?cosx+C)(6)cosxdx=d(sinx+C)

2

(7)11?x2dx=d(arcsinx+C)(8)

x1?x2dx=d(1?x2+C)(9)tanxsecxdx=d(secx+C)(10)

1x2?1dx=d(arctanx+C)(11)1x2(x?1)xdx=d(2arctanx+C)(12)xdx=d(2+C)2.求以下不定積分：

1x(1)?e5xdx?5?e5xd5x?e55?C

(2)?(2?x)72dx???(2?x)72d(2?x)??29(2?x)92?C

（3）?dx1?3x??1d(1?3x)13?1?3x??3ln|1?3x|?C（4）?exd(ex?3)ex?3dx??ex?3?ln(ex?3)?C

(5)?(e2x?2e3x?2)exdx??(e2x?2e3x?2)d(ex)?1e3x?1e4x?2ex32?C

（6）?6xd(x2?3)2x2?3dx?3?x2?3?3ln(x?3)?C

(7)

?x1x2?4dx??x2?4d(x2?42)?12?(x2?4)?12d(x2?4)

1?(x2?4)2?C?x2?4?C

（8）?x3cosx4dx?14?cosx4dx4?14sinx4?C(9)?ln4xln5xxdx??ln4xd(lnx)?5?C

1x1(10)?ex11x2dx??ed(?x)??ex?C

?3x(11)?exxdx??23?e?3xd(?3x)??23e?3?C

(12)

?arctanxx(1?x)dx?2?arctanx1?xdx?2?arctanx1?(x)2d(x)

?2?arctanxd(arctanx)?(arctanx)2?C

d(3x(13)

?dx2)4?9x2??dx?1?1321?(3x23?2)1?(3x3arcsinx2?C2)23

(14)

?arcsinx1?x2dx??arcsinxd(arcsinx)?arcsin2x2?Cx(15)

?10arccos1?x2dx???10arccosxd(arccosx)??10arccosxln10?C（16）?tanx2?1x22sinx2?1x2?1dx??tanx?1dx?1??cosx2?1dx2?1???dcosx2?1cosx2?1??ln|cosx2?1|?C

（17）?cosxsin3xdx??sin3xdsinx?14sin4x?C

(18)?sinxcosxdx???1cosxd(cosx)??2cosx?C2(19)

?sinx?cosx33sinx?cosxdx??13sinx?cosxd(sinx?cosx)?2(sinx?cosx)?C

(20)

?1?lnx(xlnx)2dx??1(xlnx)2d(xlnx)??1xlnx?C

(21)

?1xlnxlnlnxdx??1lnxlnlnxd(lnx)??1lnlnxd(lnlnx)?lnlnlnx?C

(22)?cos4xdx??(1?cos2x21?2cos2x?cos22x2)dx??4dx

??(1cos2xcos22xx?sin2x1?cos4x4?2?4)dx?4??2dx?3x?sin2xsin4x4?4?C

(23)?cos3xdx??cos2xcosxdx??(1?sin2x)d(sinx)?sinx?sin3x3?C

(24)?sin3xcos5xdx??sin2xcos5xdcosx???(1?cos2x)cos5xdcosx

?118cos8x?6cos6x?C(25)?tan3xsec5xdx?tan2xsec4xdsecx??(sec2x?1)sec4xdsecx

?17secx7x?15sec5x?C(26)?cos5xsin4xdx??sin9x?sinx2dx??1118cos9x?2cosx?C(27)?tan3xsec4xdx??tan3xsec2xdtanx??tan3x(tan2x?1)dtanx

4

?1tanx6x?1tan564x?C(28)設x?tant（|t|??2)，

?dxsec2tdtcostdtdsint11x2x2?1??tan2tsect??sin2t??sin2t??sint?C???x2x?C(29)設x?sint（|t|??2)，

?x2dx1?x2??sin2t?costdtt??sin2costdt

?12(t?tsintcost)?C?12(arcsinx?x1?x2)?C(30)令x?sect，t?[0,?2]，dx?secttantdt，則

?1xx2?1dx??1secttantsecttantdt??dt?t?C=arccos1x?C(31)令x?4sect，t?[0,?2]，dx?4secttantdt，則

2?x?16xdx??4tant4sect?4secttantdt?4?(sec2t?1)dt?4(tant?t)?C?4??x2?164???arccos??C?x2?16?4arccos41?C?3x??x（32）?dxdx1d(x?12)?1arctan(x?14x2?4x?5??(2x?1)2?4?4?)?C(x?1)2?14223d(x2?2x?17)?2d（33）?3x?1xx2?2x?17dx??2x2?2x?17

3d(x2?2??2x?17)dx321x?1x2?2x?17?2?(x?1)2?42?2ln(x?2x?17)?2arctan4?C（34）?dxdx1111x2?3x?4??(x?4)(x?1)?5?(x?4?x?1)dx?5lnx?4x?1?C

（35）?x?1x2?5x?6dx?12?d(x2?5x?6)7dxx2?5x?6?2?(x?2)(x?3)

?17?12ln(x2?5x?6)?2???x?2?1?127x?2x?3??dx?2ln(x?5x?6)?2lnx?3?C(36)?x3x2?4dx?12?x221?4?2x2?4dx?2???1?x2?4??dx

5

?12?dx2?2?d(x2?4)12x2?4?2x?2ln(x2?4)?C

(37)?x?arctan2xx2?1dx??xarctan2xx2?1dx??x2?1dx?1ln(x212?1)?3arctan3x?C

(38)

?1ex?e?xdx??1xarctanexe2x?1d(e)??C

（39）

?x1?x2?1dx??x(1?x2?1)(1?x2?1)(1?x2?1)dx??(x2?xx2?1)dx

??xx?1dx?x33??x2dx2122x3(x2?1)23?2?x?1d(x?1)?3?3?C（40）

?a?x(a?x)2a?xdx??a2?x2dx??a?xa2?x2dx?a?dxa2?x2??xa2?x2dx?aarcsinx1d(a2?x2)xa?2?a2?x2?aarcsina?a2?x2?C

1(41)

?1xx21?x2dx???1d(?1x)??d(1x(1)2?x?1(1)2?1x)x??2?1d((12121?x2x)?1)??2(x)?1?2x?C(1)2x?1（42）?sin2x1?sin2xdx?????1?1?111?sin2x??dx??dx??sin2x?dx1?1sin2xd(cotx?x??dcotx1)2?cot2x?x?22??1???cotx?2x?12arctan??cotx??2???C?2??

習題4-3

求以下不定積分（1）?xsin2xdx?12?xd(?cos2x)??x12cos2x?2?cos2xdx??x2cos2x?14sin2x?C

（2）?xe?xdx???xde?x??xe?x??e?xdx??xe?x?e?x?C

6

x3x3x3x3x2（3）?xlnxdx??lnxd()?lnx??d(lnx)?lnx??dx

2x3x3?lnx??C3333339（4）略.

（5）?x2cosxdx??x2dsinx?x2sinx??sinxdx2?x2sinx??2xsinxdx

?x2sinx??2xdcosx?x2sinx?2xcosx?2?cosxdx

?x2sinx?2xcosx?2sinx?C

(6)?arctan3x?1dx?xarctan3x?1??xdarctan3x?1?xarctan3x?1?1dx2?3x?1?xarctan3x?1?133x?1?C

（7）?e?xsin2xdx???sin2xde?x??e?xsin2x??e?xd(sin2x)

??e?xsin2x?2?cos2xd(e?x)??e?xsin2x?2e?xcos2x?2?e?xd(cos2x)??e?xsin2x?2e?xcos2x?4?e?xsin2xdx?x??e?x?esin2xdxsin2x?2e?xcos2x5?C（8）?x2arctanxdx??arctanxdx3x3x33?3arctanx??3darctanx

x31x3?3arctanx?3?1?x2dx?x33arctanx?1x3?x?x3?1?x2dx?x33arctanx?13x2?ln(1?x2)?C21?cos2x1x2（9）?xcosxdx??x2dx?2?(x?xcos2x)dx?14?2?xcos2xdx

?x24?14?xdsin2x?x24?14xsin2x?14?sin2xdx?x24?14xsin2x?18cos2x?C（10）?1xarcsinxdx?2?arcsinxdx?2xarcsinx?2?xdarcsinx

?2xarcsinx??11?xdx?2xarcsinx?21?x?C）?x2e3xdx?13?x2de3x?x2e3x3?23?xedx?x2e3x（113x3?29?xde3x

?x2e3x23x3?9xe?227e3x?C

（12）由于?coslnxdx?xcoslnx??xdcoslnx?xcoslnx??sinlnxdx

?xcoslnx?xsinlnx??xdsinlnx

7

?xcoslnx?xsinlnx??coslnxdx

于是?coslnxdx?xcoslnx?xsinlnx2?C

（13）?xcosxsin3xdx??xsin3xdsinx??xsin3xd????1?xdx2sin2x????2sin2x??sin2x??x1212sin2x?2?cscxdx??2(xcsc2x?cotx)?C（14）?ln(ex?1)exdx???ln(ex?1)de?x??e?xln(ex?1)??e?xdln(ex?1)?xx?xex??eln(e?1)??edxex?1dx??e?xln(ex?1)??ex?1ex?1?ex??e?xln(ex?1)??ex?1dx

??e?xln(ex?1)?x?ln(ex?1)?C

（15）?ln(x?1)(2?x)2dx??ln(x?1)d??1??2?x???ln(x?1)2?x??dx(2?x)(x?1)?ln(x?1)dxln(x?1)1?112?x??(2?x)(x?1)?2?x?3???2?x??x?1??dx?ln(x?1)2?x?13ln1?x2?x?C（16）?xf??(x)dx??xdf?(x)?xf?(x)??f?(x)dx?xf?(x)?f(x)?C

習題4-4

求以下不定積分

（1）?x3x?1dx??x3?1?1x?1dx??(x2?x?1)dx??1x?1dx

x3x2?3?2?x?lnx?1?C（2）?x5?x4?8x2?xx3?xdx??(x2?x?1)dx???8x3?xdx

??(x2?x?1)dx??(8x?43x?1?x?1)dx

x3?3?x22?x?8lnx?4lnx?1?3lnx?1?C2x2（3）??2x?13?x?2?3x?4(x?2)(x2?1)2dx??1x?2dx??x2?1dx??(x2?1)2dx?lnx?2?1d(x2?1)2?x2?1?2?1x?1dx?32?d(x2?1)42(x2?1)2??(x2?1)2dx?lnx?2?12ln(x2?1)?2arctanx?32x2(x2?1)?x2?1?2arctanx?C

（上式最終一個積分用積分表公式28）

8

（4）?6x2?11x?4x(x?1)2dx??[4x?2x?1?1(x?1)2]dx?4lnx?2lnx?1?1x?1?C?2lnx2(x?1)?1x?1?C（5）?xx1dx1x?x3?x2?x?1dx??(x?1)(x2?1)dx?2?x?1?12?x2?1dx?12lnx?1?14ln(x2?1)?12arctanx?C（6）?dx3?sin2x??2dx7?cos2xu?tanx?du1du3?4u2?3??(2

123u)?12tanx23arctan3?C

（7）令u?tanx2，可得2?dxsinx?cosx?1??1?u2du2u?du?ln1?u?C?ln1?tanx?C1?u2?1?u21?u2?1?u?12

（8）?dxt?33t2dt1321?31?x1?x?1?t?3?(t?1?1?t)dt?2t?t?lnt?1?C（9）?1?xx1?xdxt?1?x1?x?4t2??(112(t2?1)(t2?1)dtt?1?t?1?t2?1)dt

?lnt?1t?1?2arctant?C習題4-5

利用積分表計算以下不定積分：（1）由于?dx?2)5?4x?x2??d(x1?(x?2)2

在積分表中查得公式（73）

?dxx2?a2?ln(x?x2?a2)?C

現(xiàn)在a?1，x?x?2，于是

?dx5?4x?x2?ln(x?5?4x?x2?2)?C

（2）在積分表中查得公式（135）

?lnnxdx?x(lnx)n?n?lnn?1xdx

現(xiàn)在n?3，重復利用此公式三次，得

?ln3xdx?xln3x?3xln2x?6xlnx?6x?C.（3）在積分表中查得公式（28）

9

?(b?ax122)dx?x1dx?2?2b(ax?b)2bax2?b于是現(xiàn)在a?1，b?1，于是

?1(1?x2)2dx?x2(x2?1)?1?dx2x2?1?x2(x2?1)?arctanx?C（4）在積分表中查得公式（51）

?1xx2?adx?1aarccosax?C于是現(xiàn)在a?1，于是

?dx?arccos1xx2?1x?C（5）令t?x?1，由于

?x2x2?2xdx??x2(x?1)2?1dx??(t2?2t?1)t2?1dt

由積分表中公式（56）、（55）、（54）

?x2x2?a2dx?x2222a28(2x?a)x?a?8lnx?x2?a2?C

?xx2?a2dx?13(x2?a2)3?C?x2?a2dx?x22a2lnx?x2?a22x?a?2?C

于是

?x2x2?2xdx?x?18[2(x?1)2?a2)(x?1)2?a2?5a28lnx?1?(x?1)2?a2?13[(x?1)2?a2]3?C.（6）在積分表中查得公式（16）、（15）

?dxx2ax?b??ax?babx?2b?dxxax?b?dx2arctanax?bxax?b??b?b?C于是現(xiàn)在a?2，b??1，于是

?dx2x?1x22x?1?

2x?1dxx??x2x?1?x?2arctan2x?1?C（7）在積分表中查得公式（135）

?cosnxdx?1ncosn?1xsinx?n?1n?cosn?2xdx現(xiàn)在n?6，重復利用此公式三次，得

?cos6xdx?16cos5xsinx?53151x24?cosxsinx?24(4sin2x?2)?C.（8）在積分表中查得公式（128）

10

令

12?x?12sect?1?x1dx?12?sec2t?sectdt(x?)2?(1)222?12(tant?ln|sect?tant|)?C

?12(2x2?x?ln|2x?1?2x2?x|)?C

所以原式=x?xx2?x2?2?14ln2x?1?2x2?x?C（7）令x?tant

?lnx3dx?(1?x2)2?lntantsec3tsec2tdt??costlntantdt?sintlntant??sintsec2ttantdt?sintlntant??sectdt?sinttant?lnsect?tant?C?xlnx21?x2?lnx?1?x?C；(8)?ex(1?sinx)1?cosxdx??ex1?cosxdx+?exsinx1?cosxdx

x2exsinxcosx=?edx+22dx?exdtanx?extanxdx2cos2x?2x?2?222cos2?extanx2??extanx2dx??extanxx2dx?extan2?C；(9)?x2(1?x)100dx??(1?x)2?(1?x)?1(1?x)100dx

??[(1?x)?98?2(1?x)?99?(1?x)?100]dx

?197(1?x)?97?149(1?x)?98?199(1?x)?99?C；（10）?x21?x2arctanxdx?xarctanx?12ln(1?x2)?12(arctanx)2?Cxcos4x（11）.

?2sin3xdx??18xcsc2x2?14cotx2?C三、設x?y(x?y)2，求

?dxx?3y．

3解：令u?x?y,得x?(x?u)u2,x?uu4?3u2u2?1,dx?(u2?1)2du,16

u3?3u