企源知识库

计算下列不定积分：
(1)∫xcosxdx；
(2)∫lnxdx；
(3)∫arccosxdx；
(4)∫xe^-xdx；
(5)∫fxln(x²+1)dx；
(6)∫xsin2xdx；
(7)∫xtan²xdx；
(8)∫xe^-(x/2)dx；
(9)∫fxarcsinxdx；
(10)∫ln²xdx；
(11)∫x²sin²xdx；
(12)∫(ln³x/x²)dx；
(13)∫e^-2xcosxdx；
(14)∫ln(x+√x²+1)dx；
(15)∫e^√xdx；
(16)∫(arccosx/√1-x)dx；
(17)∫xsinxcosxdx；
(18)∫(xcosx/sin³x)dx；
(19)∫x²/(1+x²)≥arctanxdx；
(20)∫sin(lnx)dx．
【正确答案】：(1)∫xcosxdx =∫xdsinx =xsinx-∫sinxdx =xsinx+cosx+C (2)∫lnxdx=xlnx-∫xdlnx=xlnx-x+C (3)∫arccosxdx =xarccosx-∫darccosx =xarccosx+∫xdx/√1-x² =xarccosx+[-(1/2)]∫[(dl-x²)/√1-x²] =xarccosx-√1-x²+C (4)∫xe^-xdx =∫xde^-x =-xe^-x+∫e^-xdx =-xe^-x-e^-x+C (5)∫xln(x²+1)dx =1/2∫ln(x²+1)d(x²+1) =(x²+1)ln(x²+1)/2-1/2∫i(x²+1)dln(x³+1) =(x²+1)ln(x²+1)/2-1/2∫xdx =(x²+1)ln(x²+1)/2-x²/2+C (6)∫xsin2xdx =-(1/2)∫xdcos2x =-(x/2)cos2x+1/2∫cos2xdx． =-(x/2)cos2x+(1/4)sin2x+C (7)∫xtan²xdx =∫xsinxd(1/cosx) =xtanx-∫(1/cosx)dxsinx =xtanx-∫[(sinx+xcosx)/cosx]dx =xtant∫(sinx/cosx)dx-∫xdx =xtant∫+dcosx/cosx-x²/2 =xtant+ln|cosx|-x²/2+C (8)∫xe^-(x/2)dx =-2∫xde^-(x/2) =-2xe^-(x/2)+2∫e^-(x/2)dx =-2xe^-(x/2)-4e^-(x/2)+C (9)∫xarcsinxdx =1/2∫arcsinxdx² =(x²/2)arcsinx-1/2∫darcsinx =(x²/2)arcsinx-1/2∫(x²/√1-x²)dx =(x²/2)arcsinx+1/2∫(√1-x²)dx-1/2∫(1/√1-x²)dx =(x²/2)arcsinx+(1/4)arcsinx+(x/4)√1-x²-(1/2)arcsinx+C =arcsinx/4(2x²-1)+x/4√1-x²+C (10)∫ln²xdx =xln²x-∫xdln²x =xln²x-∫x•2lnx•(1/x)dx =xln²x-2∫lnxdx =xln²x-2xlnx+2∫xdlnx =xln²x-2xlnx+2x+C (11)∫x²sin²xdx =∫x²•[(1-cos2x)/2]dx =∫(x²/2)dx-∫(x²/4)dsin2x =x³/6-x²/4sin2x+∫sin2x•(x/2)dx =x³/6-(x²/4)sin2x+∫-(x/4)dcos2x =x³/6-(x²/4)sin2x-(x/4)cos2x+sin2x/8+C (12)∫(ln³x/x²)dx =∫ln³xd(1/x) =-(ln³x)/x+∫1/x•3ln²x•(1/x)dx =-(ln³x/x)-3∫ln²xd(1/x) =-(ln³x/x)-3ln²x/x+∫(6lnx/x²)dx =-(ln³x/x)-3ln²x/x-∫6lnxd(1/x) =-(ln³x/x)-3ln²x/x-6lnx/x+∫(6/x)dlnx =-(ln³x/x)-3ln²x/x-6lnx/x-6/x+C (13)∫e^-2xcosxdx =∫e^-2xdsinx =e^-2xsinx+2∫sinx•e^-2xdx =e^-2xsinx-2∫e^-2xdcosx =e^-2xsinx-2e^-2xcosx+2∫cosx•e^-2x•(-2)dx． =e^-2xsinx-2e^-xcosx-4∫e^-2xcosxdx 于是5∫e^-2xcosxx=e^-2xsinx-(2/5)e^-2xcosx+5C 所以 ∫e^-2xcosxdx=(e^-2x/5)sinx-(2/5)e^-2xcosx+C (14)∫ln(x+√x²+1)dx =xln(x+√x²+1)-∫[x/(x+√x²+1)](1+x/√x²+1)dx =xln(x+√x²+1)-1/2∫[d(x²+1)]/√x²+1 =xln(x+√x²+1)-√x²+1+C (15)∫e^√xdx =2∫√xd^√x =2√xe^√x-2∫e^√xd√x =2√xe^√x-2e^√x+C (16)∫arccosx/√1-xdx =-2∫arccosxd√1-x =-2arccosx•(√1-x)+2∫√1-x•(1/√1-x²)dx =-2arccosx•(√1-x)-2∫(1/√1+x)dx =-2arccosx•√1-x-4√1+x+C (17)∫xsinxcosxdx =∫xsinxdsinx =xsin²x-∫sinx(sinx+xcosx)dx =xsin²x-∫[(1-cos2x)/2]dx-∫xsinxcosxdx =xsin²x-x/2+sin2x/4-∫xsinxcosxdx， 于是2∫xsinxcosxdx=xsin²x-x/2+sin2x/4+2C 所以∫xsinxcosxdx=xsin²x/2-x/4+sin2x/8+C=1/8(sin2x-2xcos2x)+C (18)∫ (xcosx/sin³x)dx =-(1/2)∫xdcsc²x =-(1/2)xcsc²x+1/2∫csc²xdx =-(1/2)xcsc²x-1/2cotx+C =-(1/2)xcsc²x-(1/2)cotx+C (19)∫[x²/(1+x²)]arctanxdx =∫arctanxdx-∫[1/(1+x²)]arctanxdx =xarctanx-∫xdarctanx-∫arctanxdarctanx =xarctanx-(1/2)ln(x²+1)-(1/2)arctan²x+c (20)∫fsin(lnx)dx =xsin(lnx)-∫xdsin(1nx) =xsin(lnx)-∫cos(lnx)dx =xsin(lnx)-[xcos(lnx)-∫xdcoslnx] =xsin(lnx)-[xcos(lnx)+∫sinlnxdx] 于是2∫sin(lnx)dx=xsin(lnx)-xcos(lnx)+2C 所以∫sin(lnx)dx=(x/2)[sin(lnx)-cos(lnx)]+C