这道数学题怎么做啊(前面的不定积分符号我就不写了,用换元积分法做)1.(3-2x)^3 dx2.dx/(1-2x)3.dx/((xInx)*(In(Inx)))4.xe^-ex^2 dx5.dx/((x+1)*(x-2))

这道数学题怎么做啊(前面的不定积分符号我就不写了,用换元积分法做)1.(3-2x)^3 dx2.dx/(1-2x)3.dx/((xInx)*(In(Inx)))4.xe^-ex^2 dx5.dx/((x+1)*(x-2))
这道数学题怎么做啊(前面的不定积分符号我就不写了,用换元积分法做)
1.(3-2x)^3 dx
2.dx/(1-2x)
3.dx/((xInx)*(In(Inx)))
4.xe^-ex^2 dx
5.dx/((x+1)*(x-2))

这道数学题怎么做啊(前面的不定积分符号我就不写了,用换元积分法做)1.(3-2x)^3 dx2.dx/(1-2x)3.dx/((xInx)*(In(Inx)))4.xe^-ex^2 dx5.dx/((x+1)*(x-2))
1.∫(3-2x)^3 dx =(3-2x=y)=-(1/2)∫ydy=
=-(1/4)(y^2)+c=-(1/4)[(3-2x)^2]+c
2.∫dx/(1-2x)=(1-2x=y)=-(1/2)∫dy/y=
=-(1/2)ln|y|+c=-(1/2)ln|1-2x|+c
3.∫dx/{(xlnx)*[ln(lnx)]}=
=(lnx=y)=∫(e^y)dy/[(e^y)y(lny)=
=∫dy/ylny=(lny=u)=∫(e^u)du/(e^u)u=
=∫du/u=lnu+c=
=lnlny+c=lnlnlnx+c
4.没理解题目,该加括号的地方加上括号.
5.∫dx/[(x+1)*(x-2)]=
=-(1/3)∫dx[1/(x+1)-1/(x-2)]=
=-(1/3)[∫dx/(x+1)-∫dx/(x-2)=
=-(1/3)[ln|x+1|-ln|x-2|]+c=
=-(1/3)ln|(x+1)/(x-2)|+c

1 =(-1/2)(3-2x)^3d(3-2x)=(-1/2)(1/4)(3-2x)^4
2 =(-1/2)d(1-2x)/(1-2x)=(-1/2)ln(1-2x)
3 =xlnxd(ln(lnx))/((xlnx)*(ln(lnx)))
=d(lnx(lnx))/(ln(lnx))
=ln(ln(lnx)))
4 =(xe^-ex^2)*(-1/...

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1 =(-1/2)(3-2x)^3d(3-2x)=(-1/2)(1/4)(3-2x)^4
2 =(-1/2)d(1-2x)/(1-2x)=(-1/2)ln(1-2x)
3 =xlnxd(ln(lnx))/((xlnx)*(ln(lnx)))
=d(lnx(lnx))/(ln(lnx))
=ln(ln(lnx)))
4 =(xe^-ex^2)*(-1/2ex)d(-ex^2)
=(-1/2e)(e^(-ex^2))d(-ex^2)
=(-1/2e)e^(-ex^2)
5 =(1/3)(dx/(x-2)-dx/(x+1))
=(1/3)(d(x-2)/(x-2)-d(x+1)/(x+1))
=(1/3)(ln(x-2)-ln(x+1))
=(1/3)ln((x-2)/(x+1))

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