因为d(e^x)=e^x*dx所以原式变为{xde^x根据分部积分的公式{uv'dx={udv=uv-{vdu所以{xde^x=xe^x-{e^xdxe^x积分为e^x所以=xe^x-e^x+c这个我们确实没学过,帮不了你了,而且也不知道这个顺序具体是在哪一部分讲的

解:原式= -1/2∫x²dcos2x = -1/2(x²cos2x - ∫xdsin2x) = -1/2(x²cos2x - xsin2x + ∫sin2xdx) =[(1-2x²)cos2x + 2xsin2x]/4 + c.

你好!∫ <0,π= ∫<0,π> - x d arctan(cosx)= [- x arctan(cosx)]<0,π> + ∫<0,π>arctan(cosx) dx= π²/4 + ∫<0,π>arctan(cosx) dx 令 t=cosx ,x=arccost ∫<0,π>arctan(cosx) dx= ∫ <1,-1> arctant / -√(1-t²) dt 奇函数,积分区间对称,其值为0 故原积分= π²/4 另法:zhidao.baidu/question/177015643.html

如下

∫﹙1→∞﹚td[e^﹙-st﹚]=t[e^﹙-st﹚]|﹙1→∞﹚- ∫﹙1→∞﹚[e^﹙-st﹚]dt=【0-e^﹙-s﹚】-﹙-1/s﹚ ∫﹙1→∞﹚[e^﹙-st﹚]d﹙-st﹚=-e^﹙-s﹚+1/s*e^﹙-st﹚|﹙1→∞﹚=-e^﹙-s﹚+1/s*[0-e^﹙-s﹚]=-e^﹙-s﹚﹙1+1/s﹚

∫xe^-xdx =-∫x de^-x =-xe^-x+∫e^-xdx =-xe^-x-e^-x+C ∫x^2 cos3x dx=(1/3)∫x^2 dsin3x =(1/3)x^2sin3x-(2/3)∫xsin3x dx =(1/3)x^2sin3x+(2/9)[∫xdcos3x] =(1/3)x^2sin3x+(2/9)[xcos3x-∫cos3xdx] =(1/3)x^2sin3x+(2/9)[xcos3x-(1/3)∫cos3xd3x] =(1/3)x^2sin3x+(2/9)[xcos3x-(1/3)sin3x]+c 以下自己整理一下啊

y= ∫-sinx*e^(-x) dx=∫sinx*[-e^(-x)]dx=∫sinx* d[e^(-x)]=sinx*e^(-x) - ∫e^(-x)*d(sinx)=sinx*e^(-x) - ∫e^(-x)*cosxdx=sinx*e^(-x) + ∫cosx* [-e^(-x)]dx=sinx*e^(-x) + ∫cosx*d[e^(-x)]=sinx*.

∫ ln(1+x²) dx =x*ln(1+x²)-∫ x*d[ln(1+x²)]→分部积分法 =x*ln(1+x²)-∫ x*2x/(1+x²) dx =x*ln(1+x²)-2∫ x²/(1+x²) dx =x*ln(1+x²)-2∫ (1+x²-1)/(1+x²) dx =x*ln(1+x²)-2∫ dx+2∫ 1/(1+x²) dx =xln(1+x²)-2x+2arctanx+C

1lim(x→3,y→0)sin(xy)/tan(xy)=lim(x→3,y→0)sin(xy)'/tan(xy)'=lim(x→3,y→0)xcos(xy)/[x/(cos(xy))^2]=lim(x→3,y→0)cos(xy)=12lim(x→0,y→2) (1+sin(xy))^(y/x)=lim(x→0,y→2)(1+.

1 de^x=e^xdx (e^x)'=e^x2 可以