班级：学号：姓名：日期：月号1第四章不定积分1.判断题：（1）√（2）√（3）√（4）×（5）×2.选择题:（多项选择）（1）B（2）ABD（3）ABD（4）ABD(5)AC(6)BC)(7)ABCD3.填空题：(1)cos22xc−+，2cos2x，221ax+.(2)2222xxxeec−−−−+.(3)ln(sectan)ln(21)xx++−.(4)2ln(1)xxc+−+(5)ln(csccot)xxc−+(6)2xc+(7)13ln()41xcx−++(8)21arctanln(1)2xxxc−++4.解：21()ln1fxxcx=−+−5.解：,0()1,0xxxfxex≤⎧⎪=⎨−>⎪⎩6.解：2222,131min{,},1261,126xcxxxxdxcxxcx⎧+≤⎪⎪⎪=−+>⎨⎪⎪−++<−⎪⎩∫7.解：1()sin414Fxxx=−+8.(1)=2ln(1)xxc−−−+(2)=2121arcsin94234xxc+−+(3)=2111ln(1)arctanln1ln422xxxxc−+−−+++(4)=484441ln(1)44(1)8(1)xxxcxx+−−+++(5)=(6)=222111lnarctan814xxcx−−++(7)=611ln(4)ln244xxc−+−+(8)=3arctanarctan3xxc++(9)=tan2xxc+(10)=33arctancot62xc−+(11)=2lntan2xc+(12)=2tan22cx−++班级：学号：姓名：日期：月号2(13)()()22arctan11arctanln122xxxxeexece=−+−+−+(14)=219ln24xxc+−+(15)=()()374441611321xxxeeec+−++(16)=()21ln12xxec−++(17)=22ln11ln1xxcxxx+−+−++(18)=()2ln1ln1lnlnxxc+−−+(19)=arctantanABABxcABB⎛⎞+⎜⎟⎜⎟⎝⎠(20)=()3133sectansectanlnsectan488xxxxxxc++++(21)=()222arcsinarcsin12arcsin1222xxxxxxxcππ⎡⎤−+−−−++⎢⎥⎣⎦(22)=cxxx+−sincoscos(23)=()1535151xcx++(24)=23231arctantan332xxc⎛⎞−++⎜⎟⎝⎠(25)=211ln21xxxcx−+⎛⎞++⎜⎟−⎝⎠(26)=21xecx++9.设tan(2)nnIdxn=≥∫，证明121tan1nnnIxIn−−=−−.()()()()()()()222212tantansec1tantantan1tan1nnnnnnnIxdxxxdxxdxxdxxIn−−−−−==−=−=−−∫∫∫∫10．(1)()1lnnnxxnI−−；(2）=()()()122arcsinarcsin11nnnxxnxxnnI−−+−−−(3)=()32312222222211412433nnnnbnnbvuvuuvdxbbb−−−−+∫.
