计算下列不定积分 计算下列不定积分？
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=-(1/3)[x*^e(-3x)-∫e^(-3x)dx] =-(1/3)[x*^e(-3x)+(1/3)∫e^(-3x)d(-3x)] =-(1/3)[x*^e(-3x)+(1/3)*e^(-3x)]=……（自己化简，我没带草稿）（2）∫xcos(4x+3)dx=(1/4)∫xcos(4x+3)d(4x+3) =(1/4)∫xdsin(4x+3) =(1/4)[x*sin(4x+3)-∫sin(4x+3)dx] =(1/4)[x*sin(4x+3)-(1/4)∫sin(4x+3)d(4x+3)] =(1/4)[x*sin(4x+3)+(1/4)cos(4x+3)]=……(3))∫xsin^2 xdx=∫x*(1-cos2x)/2dx =∫0.5xdx-(1/2)∫xcos(2x)dx =(x^2)/4-(1/4)∫xcos(2x)d(2x) =(x^2)/4-(1/4)∫xd(sin2x) =(x^2)/4-(1/4)[xsin(2x)-∫sin(2x)dx] =(x^2)/4-(1/4)[xsin(2x)-(1/2)∫sin(2x)d2x] ==(x^2)/4-(1/4)[xsin(2x)+(1/2)cos2x]=……(4)∫x^2 lnxdx=(1/3)∫(3x^2)lnxdx =(1/3)∫lnxd(x^3) =(1/3)[x^3*lnx-∫x^3d(lnx)] =(1/3)[x^3*lnx-∫x^2dx]
=(1/3)[x^3*lnx-(1/3)x^3]=……(5)设√x=t ∫cos√xdx=∫costdt^2 =2∫tcostdt =2∫td(sint) =2[t*sint-∫sintdt] =2[t*sint+cost] =2[√x*sin(√x)+cos(√x)]=……所有的结果别忘了加常数C吖
都是分部积分的练习：(1)∫xe^(-3x)dx=-1/3∫xe^(-3x)d(-3x)=-1/3∫xd[e^(-3x)]=-1/3[xe^(-3x)-∫e^(-3x)dx]=
(2)∫xcos(4x+3)dx=1/4∫xcos(4x+3)d(4x+3)=1/4∫xd[sin(4x+3)]=1/4xsin(4x+3)-1/4∫sin(4x+3)dx=
(3)sin^2x=(1-cos2x)/2,∴∫xsin^2 xdx=1/2∫xdx-1/4∫xcos2xd(2x),前一个是简单积分，后一个仍是分部积分与（2）类似。请练习。
(4)∫x^2 lnxdx=∫ lnxd(x�0�6/3)=x�0�6lnx/3-∫ (x�0�6/3)d(lnx)=x�0�6lnx/3-∫ (x�0�5/3)dx=x�0�6lnx/3-x�0�6/9+C(5)求∫cos√xdx,此题要同时运用换元积分与分部积分，换元u=√x,x=u�0�5,dx=2udu∫cos√xdx=2∫ucosudu=...后面的分部积分各书上都有例题可参照，就不代替了，数学是一定要自己多练习的，所以以上各题遵照你的意思写明了详细解题过程，但最终完成请自己做吧！
绛旓細绗竴棰 鏈涢噰绾
绛旓細瑙ｏ細锛1锛夆埆xe^(-3x)dx=-(1/3)鈭玿e^(-3x)d(-3x)=-(1/3)鈭玿de^(-3x)=-(1/3)[x*^e(-3x)-鈭玡^(-3x)dx] =-(1/3)[x*^e(-3x)+(1/3)鈭玡^(-3x)d(-3x)] =-(1/3)[x*^e(-3x)+(1/3
绛旓細濡傚浘鎵绀
绛旓細鍒欙細x=(lnt)/2锛宒x=1/(2t)dt 鈭玠x/[1+e^(2x)]= (1/2)鈭1/[t(1+t)]dt = (1/2)鈭玔(1+t)-t]/[t(1+t)]dt = (1/2)鈭玔1/t - 1/(1+t)]dt = (1/2)[ln|t|-ln|1+t|]+C = (1/2)[ln|e^(2x)| - ln|1+e^(2x)]+C = x-(1/2)ln|1+e^(2x)|...
绛旓細1銆佸師寮=鈭玠(1+e^x)/(1+e^x)=ln|1+e^x|+C锛屽叾涓瑿鏄换鎰忓父鏁 2銆佸師寮=-鈭玸in(1/x)d(1/x)=cos(1/x)+C锛屽叾涓瑿鏄换鎰忓父鏁 3銆佸師寮=2鈭玸in鈭歺d(鈭歺)=-2cos鈭歺+C锛屽叾涓瑿鏄换鎰忓父鏁
绛旓細1銆佲埆 (x+2)dx/鈭(x²-2x+4)= (1/2)鈭 (2x-2+6)dx/鈭(x²-2x+4)= (1/2)鈭 (2x-2)dx/鈭(x²-2x+4) + 3鈭 dx/鈭(x²-2x+4)= (1/2)鈭 d(x²-2x+4)/鈭(x²-2x+4) + 3鈭 dx/鈭歔(x-1)²+3]= (1/2)*2鈭(x&...
绛旓細鈭玹anxdx =鈭玸inx/cosx dx =鈭1/cosx d(-cosx),娉ㄦ剰鈭玸inxdx=-cosx,鎵浠inxdx=d(-cosx) =-鈭1/cosx d(cosx),浠=cosx,du=d(cosx) =-鈭1/u du =-ln|u|+C =-ln|cosx|+C 鎵╁睍璧勬枡鍦ㄥ井绉垎涓紝涓涓嚱鏁癴 鐨涓嶅畾绉垎锛屾垨鍘熷嚱鏁锛屾垨鍙嶅鏁帮紝鏄竴涓鏁扮瓑浜巉 鐨勫嚱鏁 F 锛屽嵆F...
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