4-1高等数学同济大学第六版本

习题411.求下列不定积分:(1)dxx21;解CxCxdxxdxx112111222.(2)dxxx;解CxxCxdxxdxxx212323521231.(3)dxx1;解CxCxdxxdxx21211112121.(4)dxxx32;解CxxCxdxxdxxx3313737321031371.(5)dxxx21;解CxxCxdxxdxxx12312511125252.(6)dxxmn;解CxmnmCxmndxxdxxmnmmnmnmn111.(7)dxx35;解Cxdxxdxx4334555.(8)dxxx)23(2;解Cxxxdxdxxdxxdxxx2233123)23(2322.(9)ghdh2(g是常数);解CghChgdhhgghdh22212122121.(10)dxx2)2(;解Cxxxdxdxxdxxdxxxdxx423144)44()2(23222.(11)dxx22)1(;解Cxxxdxdxxdxxdxxxdxx3524242232512)12()1(.(12)dxxx)1)(1(3;解dxdxxdxxdxxdxxxxdxxx23212323)1()1)(1(Cxxxx25233523231.(13)dxxx2)1(;解Cxxxdxxxxdxxxxdxxx2523212321212252342)2(21)1(.(14)dxxxx1133224;解Cxxdxxxdxxxxarctan)113(1133322224.(15)dxxx221;解Cxxdxxdxxxdxxxarctan)111(111122222.(16)dxxex)32(;解Cxedxxdxedxxexxx||ln32132)32(.(17)dxxx)1213(22;解Cxxdxxdxxdxxxarcsin2arctan3112113)1213(2222.(18)dxxeexx)1(;解Cxedxxedxxeexxxx21212)()1(.(19)dxexx3;解CeCeedxedxexxxxxx13ln3)3ln()3()3(3.(20)dxxxx32532;解CxCxdxdxxxxxxx)32(3ln2ln5232ln)32(52])32(52[32532.(21)dxxxx)tan(secsec;解Cxxdxxxxdxxxxsectan)tansec(sec)tan(secsec2.(22)dxx2cos2;解Cxxdxxdxxdxx)sin(21)cos1(212cos12cos2.(23)dxx2cos11;解Cxdxxdxxtan21cos212cos112.(24)dxxxxsincos2cos;解Cxxdxxxdxxxxxdxxxxcossin)sin(cossincossincossincos2cos22.(25)dxxxx22sincos2cos;解Cxxdxxxdxxxxxdxxxxtancot)cos1sin1(sincossincossincos2cos22222222.(26)dxxxx)11(2;解dxxxx211Cxxdxxx41474543474)(.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3213f(e2)ln|e2|C2C,C32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xxexshch的原函数.证明xxxxxxxxxeeeeeeeexxe222shch.因为xxee22)21(,所以xe221是xxexshch的原函数.因为(exshx)exshxexchxex(shxchx)xxxxxxeeeeee2)22(,所以exshx是xxexshch的原函数.因为(exchx)exchxexshxex(chxshx)xxxxxxeeeeee2)22(,所以exchx是xxexshch的原函数.