√1-X的平方dx=4 π-搜答案-智慧作业帮

令x=r*sinz,dx=r*coszdz√(r²-x²)=√(r²-r²*sin²z)=r*coszcosz=√(r²-x²)/

[x√(1-y²)]dx+[y√(1-x²)]dy=0[y√(1-x²)]dy＝-[x√(1-y²)]dx分离变量得ydy/√(1-y²)=-xdx/

∫dx/(x²+2x+5)=∫d(x+1)/[(x+1)²+4]=(1/2)arctan[(x+1)/2]+C

1.∫1/cos(x)dx=∫cosx*dx/(cos(x))^2=∫d(sinx)/(1-(sinx)^2)=-1/2*∫2*d(sinx)/((sinx-1)*(sinx+1)=-1/2*((∫d

1/[x^2(1-x)]=A/(x-1)+B/x+C/x^2Ax^2+Bx(x-1)+C(x-1)=-1x=0:C=1x=1:A=-1x=-1:A+2B-2C=-1B=1∫dx/[x^2(1-x)]=

x(1-y^2)^(1/2)dx+y(1-x^2)^(1/2)dy=0,|x|

∫(x^2+√x)dx=(1/3)x^3+2x√x/3+C

见附图.

y=1-1/(1+x),一阶导数是1/(1+x)^2,二阶导数是-2/(1+x)^3

你应该说的是∫(x^x)(1+lnx)dx=∫[e^(xlnx)](1+lnx)dx=∫[e^(xlnx)]d(xlnx)=e^(xlnx)+c=x^x+c

再答：复合函数求导再答：-x²求导不是-2x嘛，题目中没有-2，所以乘以一个-1/2再问：不是求积吗为什么要对-x^2求导？(>_

点击看大图

∫（x-1）^2dx=∫(x^2-2x+1)dx=1/3x^2-x^2+x+C

x/Sqrt[1+x^2]+ln(x+Sqrt[1+x^2])

∫1/[x√(1-x²)]dx=∫1/[x*√[x²(1/x²-1)]dx=∫1/[x*|x|*√(1/x²-1)]dx=∫1/[x²√(1/x

=∫(x^2-4x+4)*x^(-1/2)dx=∫[x^(3/2)-4x^(1/2)+4x^(-1/2)]dx=2x^(5/2)/5-8x^(3/2)/3+8x^(1/2)+C=2x^2√x-8x√x

因为d(1/x)=-1/x^2dx所以∫{cos(1/x)/x^2}dx=(-1)*∫cos(1/x)d(1/x)=-sin(1/x)+C