—1—0基础知识06课后小测验答案1.设函数()fx在(),上连续,则[()]dfxdx()A.()fxB.()fxdxC.()fxCD.()fxdx答案:B解析:()()()dfxdxfxdxdxfxdx2.(1)xxeedxA.xexB.xexCC.xexCD.xexC答案:B解析:(1)(1)xxxxeedxedxexC3.2211xdxxA.2arctanxxB.2arctanxxC.2arctanxxCD.2arctanxxC答案:D解析:222221122(1)1112arctanxxdxdxdxxxxxxC4.1(1)dxxxA.2arctanxCB.2arctanxC.2arctanxxCD.2arctanxC—2—答案:A解析:令xt,则2xt,22dxdttdt,22211122(1)(1)(1)22arctan2arctan1dxtdttdtttttxxdttCxCt5.21xdxA.211arcsin22xxxB.211arcsin22xxxC.211arcsin22xxxCD.211arcsin22xxxC答案:C解析:令sinxt,则sincosdxdttdt,22221cos211sincoscos21111sin2arcsin2422txdxttdttdtdtxxttCxC其中回代过程中2sin22sincos21tttxx6.211dxxA.2ln1xxB.2ln1xxCC.2ln1xxCD.2ln1xxC—3—答案:B解析:令tanxt,则21secxt,2tansecdxdttdt,22211secsecsec1lnsectanln1dxtdttdttxttCxxC7.arctanxxdxA.211arctanarctan222xxxxB.211arctanarctan222xxxxC.211arctanarctan222xxxxCD.211arctanarctan222xxxxC答案:D解析:使用分部积分法222222222222arctanarctanarctanarctan2221111arctanarctan2212211111arctan(1)arctanarctan221222xxxxxdxxdxdxxxxxxdxxdxxxxxxdxxxxCx8.2123dxxxA.1ln3xCxB.11ln43xCxC.11ln43xCxD.1ln3xCx—4—答案:C解析:211111()23(1)(3)413111(ln1ln3)ln443dxdxdxxxxxxxxxxCCx