陈文灯主编《微积分》第四章作业解答

第四章作业解答

1．用积分公式直接求下列不定积分：

（1）?9x4?4xx?1x3dx

9x4?4xx?133解：?x3dx??(9x?4x?2?x?3)dx?9?xdx?4?x?2dx??x?3dx 91?2?2?x9812x?8x2?2?C?2x2?x?2x2?C （3）?3x4?3x2?1x2?1dx 解： ?3x4?3x2?1x2?1dx??3x2(x2?1)?1x2?1dx??3x2dx??1x2?1dx ?x3?arctanx?C

（5）?(3e)5xdx

[(3e)5]x解：?(3e)5xdx??[(3e)5]xdx?(3e)5xln(3e)5?C?5ln3?5?C （7）?sin2x2dx

解： ?sin2x1?cosx2dx??2dx?12?dx?12?cosxdx?12x?12sinx?C 2．用积分公式直接求下列不定积分：

（1）?(x?1)2xdx

解：?(x?1)231xdx??x2?2x?1xdx??(x2?2x2?x?12)dx 531?2x2?4x2?2x253?C （2）?xxxdx 1715解：?xxxdx??2?1x4?18dx??x8dx?8815x?C

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（3）?cos2xdx

sin2xcos2xcos2x11cos2x?sin2xdx?(??dx解：?2?sin2xcos2x)dx ?sin2xcos2xsinxcos2x??(csc2x?sec2x)dx??cotx?tanx?C

1?cos2xdx （4）?1?cos2x111111?cos2x1?cos2x?dx?dx?tanx?x?C dx??dx解：?2??22cosx2221?cos2x2cosxex(x?e?x)dx （5）?x1ex(x?e?x)dx??exdx??dx?ex?lnx?C 解：?xx3．用第一类换元法求下列不定积分： （1）?e?xdx

解：?e?xdx???e?xd(?x)??e?x?C （2）?42x?1dx 解：?41(2x?1)12?C?(2x?1)4?C 2x?1dx??(2x?1)d(2x?1)?5225414545（3）?解：?dx

(1?2x)311dx11?3?2??(1?2x)d(1?2x)?(1?2x)?C ??d(1?2x)?33?242(1?2x)(1?2x)（4）?axexdx

(ae)xaxex?C??C 解：?aedx??(ae)dx?lnaelna?1xxx（5）?dx

4?9x2 2

解：?dx1dx4?9x2?4??1?21?(3x243?13x13x3xd()?arctan?C 22)1?(2)262（6）?dxcos2(2x??

4)解：?dx?1cos2(2x??4)2?1d(2x??1?cos2(2x??4)?2tan(2x?4)?C

4)（7）?sec2xtanxdx 解：?sec2x1tanxdx??tanxdtanx?lntanx?C （8）?xdx1?2x2

解：?xdx1?2x2?14?11?2x2d(1?2x2)?112(1?2x2)2?C

4．用第一类换元法求下列不定积分： （1）?x22?x3dx

2解：?xdx112?x3??3?2?x3d(2?x3)??13?(2?x3)?12d(2?x3) ??2123(2?x3)2?C??2?x33?C （2）?dxex?e?x 解：?dxexdxex?e?x??e2x?1??1e2x?1dex?arctanex?C （3）?dxx2?2x?5 解：?dxdxx2?2x?5??dxdx1x2?2x?1?4??(x?1)2?4?4?1?(x?1 22)?11x?x?14?2?d(1)?1arctan?C 1?(x?122222)

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（4）?1x?x2dx

解：?111x?x2dx??x1?xdx?2?dx?2arcsinx?C

1?(x)2arctan1（5）?x1?x2dx arctan1解：?x1?x2dx???arctan1xdarctan1x??12(arctan1x)2?C （10）?tan3xdx

解：?tan3xdx=?tanx(sec2x?1)dx??tanxsec2xdx??tanxdx

??tanxdtanx??tanxdx?12tan2x?lncosx?C （11）?sin3xdx

解：?sin3xdx??sinxsin2xdx???(1?cos2x)dcosx

???dcosx??cos2xdcosx??cosx?13cos3x?C

5．用第二类换元法求下列不定积分：

（1）?x3?xdx

解：设t?3?x，则x?t2?3，dx?2tdt 所以?x3?xdx??(t2?3)?t?2tdt??(t2?3)?t?2tdt

53?2?(t4?3t2)dt=?225t5?2t3?C?5(3?x)2?2(3?x)2?C

（3）?dxx?3x

解：设t?6x，则x?t6，dx?6t5dt

dx3?6t5dttt3?1?1(t?1)(t2x?3x??t3?t2?6?t?1dt?6?t?1dt?6?(?t?1)t?1?1t?1)dt?6?(t2?t?1)dt?6?1t?1dt?2t3?3t2?6t?6lnt?1?C 4

?2x?33x?66x?6ln6x?1?C

（5）?dx1?e2x

t1解：设t?1?e2x，则x?2ln(t2?1)，dx?t2?1dt t所以?dx??t2?1tdt??11111?e2xt2?1dt?2?(t?1?t?1)dt ?111?e2x?11e2x2(lnt?1?lnt?1)?C?2ln1?e2x?1?C?2ln(1?e2x?1)2?C ?1lne2x?1ln(1?e2x?1)2??C?x?ln(1?e2x22?1)?C （8）?x34?x2dx

解：设x?2sint，则dx?2costdt

所以?x34?x2dx??8sin3t?4?4sin2t?2costdt

?32?sin3t?cos2tdt??32?(cos2t?cos4t)dcost??32(cos3tcos5t3?5)?C35??323(4?x22)3?325(4?x22)5?C??413(4?x)2?5(4?x)2?C

（10）?dxx2(1?x2)3

解：设x?sint，则dx?costdt 所以?dxcostdtx(1?x2)3??sin2t(1?sin2t)3??costdtsin2tcos3t??sin2t?cos2t2sin2tcos2tdt??(sec2t?csc2t)dt?tant?cost?C?x1?x21?x2?x?C 2（12）?xdx(1?x2)3

解：设x?tant，则dx?sec2tdt

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所以?x2dx(1?x)23??tan2tsec2tdttan2tsec2tdt?? 323sect(1?tant)??(sect?cost)dt?lnsect?tant?sint?C?ln1?x2?x?x?C

1?x26．用分部积分法求下列不定积分： （1）?(ex?lnx)dx

解：?(ex?lnx)dx??exdx??lnxdx?ex?xlnx??xdlnx

?ex?xlnx??dx?ex?xlnx?x?C

（2） ?x2e?2xdx 解：?x2e?2xdx??12?x2de?2x??12x2e?2x?12?e?2xdx2 ??1x2e?2x2??xe?2xdx??1x2e?2x?12?xde?2x2

??1x2e?2x2?1xe?2x?1?e?2xdx??1x2e?2x?1xe?2x12222?4?e?2xd(?2x) ??1x2e?2x2x2?12xe?2x?14e??C??112e?2x(x2?x?2)?C （3）?(2x?3x2)arctanxdx 解：?(2x?3x2)arctanxdx

??arctanxd(x2?x3)?(x2?x3)arctanx??(x2?x3)darctanx

?(x2?x3)arctanx??x2?x31?x2dx?(x2?x3)arctanx??x(x2?1)?x2?1?x?11?x2dx ?(x2?x3)arctanx??xdx??dx??x1?x2dx??11?x2dx?(x2?x3)arctanx?x212?x?2?121?x2d(1?x)?arctanx ?(x2?x3?1)arctanx?x22?x?12ln(1?x2)?C

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（4） ?(lnx2x)dx 解：?(lnx2x??(lnx)2)dxx2dx???(lnx)2d1x??121x(lnx)??xd(lnx)2 ??1x(lnx)2??1x2lnx?111xdx??x(lnx)2?2?lnx?dx

??1x(lnx)2?211111xlnx?2?x?dlnx??x(lnx)2?2xlnx?2?x2?dx

??1(lnx)2?21lnx?21?C??1[(lnx)2xxxx?2lnx?2]?C

（5） ?sinxex

解：?sinxex??e?xsinxdx???e?xdcosx??e?xcosx??cosxde?x

??e?xcosx??e?xcosxdx??e?xcosx??e?xdsinx

??e?xcosx?e?xsinx??sinxde?x??e?xcosx?e?xsinx??e?xsinxdx所以?sinxex??e?xsinxdx??12e?x(cosx?sinx)?C （6） ?xsec2xdx

解：?xsec2xdx??xdtanx?xtanx??tanxdx

?xtanx??sinxcosxdx?xtanx??1cosxdcosx ?xtanx?lncosx?C （7） ?xtan2xdx

解： ?xtan2xdx??x(sec2x?1)dx??xsec2xdx??xdx??xdtanx?12x2 ?xtanx??tanxdx?12x2?xtanx?lncosx?12x2?C （8） ?ln(1?x2)dx

解：?ln(1?x2)dx?xln(1?x2)??xdln(1?x2)

2x22(1?x2?xln(1?x2)??1?x2dx?xln(1?x2)??)?21?x2dx ?xln(1?x2)??(2?21?x2)dx?xln(1?x2)?2x?2arctanx?C

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