nxe2Iz4•—「1X>/I+InX1d(l+lnx)1Jl+ln兀=2Jl+ln兀6=2(73-1)112.2xcos2Adv01解:2xcos2xdx=—xsin2xJo?221]sin2Adr=—cos2x413.[ln(x4-l)dxJoJ()ln(x+\)dx=xln(x+1)jc-lX——dx=e-l-x+1Cc-lL-=e-1-[x-ln(x+l)]|o1=lne=l_1-10_11.设矩阵A=-121,B=2223__5_四、代数计算题求4"B.-51-10100_1-10100_—-121010->011110223001043-201_1-10100_011110_100-4■_00-1-6-41■T010-5z1-1010o-00164■»010-5-3100164-1解:因为11-4即A」_-4-31~~r-5所以A-lB=-5-312——-664-15_964-10?-1-32.设矩阵A,/是3阶单位矩阵,求(八A)-3-4_100~/-A=010—001解:由矩阵减法运算得0?-1-3_1?13-2-2-7=237-3-4-8349_113100__113100_237010011--210349001010--301j13100-「110-2-33__1001?-3->011-210->010-301010-3?000-1-1-11_00111-100111利用初等行变换得21-11即3._63「「10-2,B=121-204111计算(AB)1.设矩阵A二「63'「10-212—_-21_1-204-141解因为AB=4•解矩阵方程-23-34-1x=2-3X4-1-23-12-23-3140「111r1043_->T01-3101-3-33_-2110__-2110_T_-20-14?-1010121_0121_10(AB/)=01所以,
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