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1、第四章习题 411. 求下列不定积分:(1)dx x21;解CxCxdxxdxx+=+=+1 1211122 2.(2)dxxx;解CxxCxdxxdxxx+=+ +=+2123 23521231.(3)dx x1;解CxCxdxxdx x+=+ +=+2 12111121 21 .(4)dxxx32;解CxxCxdxxdxxx+=+ +=+33137 37 32 1031371.(5)dx xx21;解C xxCxdxxdx xx+=+ +=+1 2312511125 252.(6)dxxmn;解CxmnmCxmndxxdxxmnm mn mn mn+=+ +=+111.(7)dxx35;解
2、Cxdxxdxx+=433 4555.(8)+dxxx)23(2;解Cxxxdxdxxdxxdxxx+=+=+223 3123)23(2322.(9) ghdh2(g是常数);解CghCh gdhh gghdh+=+=22 2121221 21 .(10)dxx2)2(;解Cxxxdxdxxdxxdxxxdxx+=+=+=423144)44()2(23222.(11)+dxx22) 1(;解Cxxxdxdxxdxxdxxxdxx+=+=+=+35242422 32 512) 12() 1(.(12)dxxx+) 1)(1(3;解+=+=+dxdxxdxxdxxdxxxxdxxx23 21 23
3、23) 1() 1)(1(Cxxxx+=25 23 3 52 32 31.(13)dx xx2)1 (;解Cxxxdxxxxdx xxxdx xx+=+=+=25 23 21 23 21 212252 342)2(21)1 (.(14)+dxxxx 1133224 ;解Cxxdx xxdxxxx+=+=+arctan)113(11333 22 224 .(15)+dxxx221;解+=+=+=+Cxxdxxdxxxdxxxarctan)111 (111122222 .(16)+dxxex)32(;解Cxedxxdxedxxexxx+=+=+|ln32132)32(.(17) +dx xx) 1
4、213( 22;解+= += +Cxxdx xdxxdx xxarcsin2arctan3 112113) 1213( 2222.(18)dx xeex x )1 (;解Cxedxxedx xeexxx x+= 21 21 2)()1 (.(19)dxexx3;解CeCeedxedxexxx xxx+=+=13ln3 )3ln()3()3(3.(20)dxxxx32532;解CxCxdxdxxxx xxx +=+=)32(3ln2ln5232ln)32( 52)32(5232532.(21)dxxxx)tan(secsec;解+=Cxxdxxxxdxxxxsectan)tansec(sec)t
5、an(secsec2.(22)dxx 2cos2;解Cxxdxxdxxdxx+=+=+=)sin(21)cos1 (21 2cos1 2cos2.(23)+dxx2cos11;解+=+Cxdxxdxxtan21cos21 2cos112.(24)dxxxx sincos2cos;解+=+=Cxxdxxxdxxxxxdxxxxcossin)sin(cossincossincos sincos2cos22 .(25)dxxxx22sincos2cos;解+=Cxxdxxxdxxxxxdxxxxtancot)cos1 sin1(sincossincos sincos2cos22222222.(26)
6、dxxx x)11 (2;解dxxxx211+=Cxxdxxx41 47 45 43 474)(.2. 一曲线通过点(e2, 3), 且在任一点处的切线的斜率等于该点横坐标的倒数, 求该曲 线的方程. 解 设该曲线的方程为y=f(x), 则由题意得xxfy1)(=,所以Cxdxxy+=|ln1.又因为曲线通过点(e2, 3), 所以有=32=13=f(e2)=ln|e2|+C=2+C, C=32=1. 于是所求曲线的方程为y=ln|x|+1. 3. 一物体由静止开始运动, 经t秒后的速度是 3t2(m/s), 问 (1)在 3 秒后物体离开出发点的距离是多少？ (2)物体走完 360m 需要多
7、少时间？解 设位移函数为s=s(t), 则s=v=3t2,Ctdtts+=323.因为当t=0 时,s=0, 所以C=0. 因此位移函数为s=t3.(1)在 3 秒后物体离开出发点的距离是s=s(3)=33=27.(2)由t3=360, 得物体走完 360m 所需的时间11. 73603=ts.4. 证明函数xe221,exshx和exchx都是xxex shch 的原函数.证明x xxxxxxxxeee eeeee xxe222shch=+=.因为xxee22)21(=, 所以xe221是xxex shch 的原函数.因为 (exshx)=exshx+exchx=ex(shx+chx)xxx
8、xxxeeeeee2)22(=+=,所以exshx是xxex shch 的原函数.因为 (exchx)=exchx+exshx=ex(chx+shx)xxxxxxeeeeee2)22(=+=,所以exchx是xxex shch 的原函数.习题 421. 在下列各式等号右端的空白处填入适当的系数, 使等式成立(例如: )74(41+=xddx:(1)dx= d(ax);解dx= a1d(ax).(2)dx=d(7x3);解dx=71d(7x3).(3)xdx=d(x2);解xdx=21d(x2).(4)xdx=d(5x2);解xdx=101d(5x2).(5)1 ( 2xdxdx=;解)1 (
9、212xdxdx=.(6)x3dx=d(3x42);解x3dx=121d(3x42).(7)e2xdx=d(e2x);解e2xdx=21d(e2x).(8)1 ( 22xxeddxe+=;解)1 ( 2 22xxeddxe+=.(9)23(cos 23sinxdxdx=;解)23(cos 3223sinxdxdx=.(10)|)|ln5( xdxdx=;解|)|ln5( 51xdxdx=.(11)|)|ln53( xdxdx=;解|)|ln53( 51xdxdx=.(12)3(arctan 912xdxdx=+;解)3(arctan 31912xdxdx=+.(13)arctan1 ( 12x
10、d xdx= ;解)arctan1 ( ) 1( 12xd xdx= .(14)1( 122xd xxdx= .解)1( ) 1( 122xd xxdx= .2. 求下列不定积分(其中a,b,均为常数):(1)dtet5;解Cexdedtexxt+=555 51551.(2)dxx3)23(;解Cxxdxdxx+=433)23(81)23()23(21)23(.(3)dxx211;解Cxxdxdxx+=|21|ln21)21 (211 21 211.(4) 332xdx;解CxCxxdx xdx+=+= 32 32 313)32(21)32(23 31)32()32(3132.(5)dxeax
11、bx )(sin;解Cbeaxabxdebaxdaxadxeaxbx bx bx +=cos1)()(sin1)(sin.(6)dt ttsin;解+=Cttdtdt ttcos2sin2sin.(7)xdxx210sectan;解xdxx210sectanCxxxd+=1110tan111tantan.(8)xxxdx lnlnln;解Cxxdxxdxxxxxdx+=|lnln|lnlnlnlnln1lnlnlnln1 lnlnln.(9) +dx xxx 2211tan;解+dx xxx 2211tan222 221 1cos1sin11tanxd xxxdx+ +=+=Cxxd x+=+
12、 +=|1cos|ln1cos 1cos1222.(10)xxdx cossin;解Cxxdxdxxx xxdx+=|tan|lntantan1 tansec cossin2 .(11)+dxeexx1;解+dxeexx1Cedeedxeexx xxx +=+=+=arctan11122.(12)dxxex2;解.21)(212222Cexdedxxexxx+=(13)dxxx)cos(2;解Cxxdxdxxx+=)sin(21)()cos(21)cos(2222.(14) dx xx232;解CxCxxdxdx xx+=+= 221 2221 223231)32(31)32()32(6132
13、.(15)dxxx4313;解+=Cxxdxdxxx|1|ln43)1 (11 431344 443 .(16)+dttt)sin(cos2;解Cttdtdttt+=+=+)(cos31)cos()(cos1)sin()(cos322.(17)dxxx3cossin;解CxCxxxddx xx+=+=223 3sec21cos21coscoscossin.(18) +dx xxxx3cossincossin;解)sincos( cossin1cossincossin33xxd xxdx xxxx+ = +Cxxxxdxx+=32 31 )cos(sin23)cos(sin)cos(sin.(1
14、9) dx xx2491;解dx xxdx xdx xx = 22249491491)49( 491 81)32()32(11 21222xd xxdx +=Cxx+=24941 32arcsin21.(20)+dxxx239;解Cxxxdxxdxxdxxx+=+=+=+)9ln(921)()991 (21)(9219222 22 2223 .(21)dxx1212;解+ = +=dx xxdx xxdxx) 121121(21) 12)(12(11212+ + =) 12( 121221) 12( 121221xd xxd xC xxCxx+ +=+=| 1212|ln 221|12|ln 221|12|ln 221.(22)+dxxx)2)(1(1;解CxxCxxdxxxdxxx+=+=+=+|12|ln31|1|ln|2|(ln31)11 21(31 )2)(1(1.(23)xdx3cos;解Cxxxdxxdxxdx+=3223sin31sinsin)sin1 (sincoscos.(24)+dtt)(cos2;解Cttdttdtt+=+=+)(2sin41 21)(2cos1 21)
