∫1/(x^3(1+x^3))dx

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∫1/(x^3+x) dx

∫1/(x^3+x)dx∫1/(x^3+x)dx∫1/(x^3+x)dx先将被积函数分解部分分式,再积分∫1/(x^3+x)dx=∫(x^2+1-x^2)/x(x^2+1)dx=∫[1/x-x/(x^

x-9/[(根号)x]+3 dx ∫ x+1/[(根号)x] dx ∫ [(3-x^2)]^2 dx

x-9/[(根号)x]+3dx∫x+1/[(根号)x]dx∫[(3-x^2)]^2dxx-9/[(根号)x]+3dx∫x+1/[(根号)x]dx∫[(3-x^2)]^2dxx-9/[(根号)x]+3d

∫1/x(1+x^3)dx

∫1/x(1+x^3)dx∫1/x(1+x^3)dx∫1/x(1+x^3)dx上下乘以X^2再积分

∫x+1/(x-1)^3dx

∫x+1/(x-1)^3dx∫x+1/(x-1)^3dx∫x+1/(x-1)^3dxD

求∫x/(1+x)^3 dx

求∫x/(1+x)^3dx求∫x/(1+x)^3dx求∫x/(1+x)^3dx

∫ x/ (1+x)^3 dx

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

∫dx/x(x^3+1)

∫dx/x(x^3+1)∫dx/x(x^3+1)∫dx/x(x^3+1)∫dx/[x(x^3+1)]=∫dx/[x(x+1)(x^2-x+1)]let1/[x(x^3+1)]≡A/x+B/(x+1)+

∫x^3/1+x^2 dx

∫x^3/1+x^2dx∫x^3/1+x^2dx∫x^3/1+x^2dx∫x^3/(1+x^2)dx=∫[x^3+x-x]/(1+x^2)dx=∫x-x/(1+x^2)dx=x²/2-1/2

∫(x-1)^2/x^3 dx

∫(x-1)^2/x^3dx∫(x-1)^2/x^3dx∫(x-1)^2/x^3dx∫(x²-2x+1)/x³dx=∫(1/x-2/x²+1/x³)dx=lnx

∫(X^3)/(1+X^2)dx

∫(X^3)/(1+X^2)dx∫(X^3)/(1+X^2)dx∫(X^3)/(1+X^2)dx具体见图片内容:

∫1/(x^3+1)dx

∫1/(x^3+1)dx∫1/(x^3+1)dx∫1/(x^3+1)dx题目不难,做起来比较繁琐一点,解答如下:

∫(1-sin^3x)dx

∫(1-sin^3x)dx∫(1-sin^3x)dx∫(1-sin^3x)dx原式=∫1dx-∫sin²x*sinxdx=x+∫(1-cos²x)dcosx=x+cosx-cos&

∫x^3/(x^8-2) dx∫(x^3-1)/(x^2+1) dx

∫x^3/(x^8-2)dx∫(x^3-1)/(x^2+1)dx∫x^3/(x^8-2)dx∫(x^3-1)/(x^2+1)dx∫x^3/(x^8-2)dx∫(x^3-1)/(x^2+1)dx1.令t

∫1/(x^100+x)dx ∫1/(e^x+e^3x)dx

∫1/(x^100+x)dx∫1/(e^x+e^3x)dx∫1/(x^100+x)dx∫1/(e^x+e^3x)dx∫1/(x^100+x)dx∫1/(e^x+e^3x)dx1、∫1/(x^100+x

∫1/(X+1)(X+3) dx.

∫1/(X+1)(X+3)dx.∫1/(X+1)(X+3)dx.∫1/(X+1)(X+3)dx.∫dx/[(x+1)(x+3)]=(1/2)∫[(x+3)-(x+1)]/[(x+1)(x+3)]dx=

∫(1,0)(x^3+√x)dx

∫(1,0)(x^3+√x)dx∫(1,0)(x^3+√x)dx ∫(1,0)(x^3+√x)dx∫(0,1)(x^3+√x)dx=[(1/4)x^4+(2/3)x^(3/2)]I(0,1)

∫x^3/(x+1)dx求不定积分

∫x^3/(x+1)dx求不定积分∫x^3/(x+1)dx求不定积分∫x^3/(x+1)dx求不定积分

∫x●dx/(1-x)^3

∫x●dx/(1-x)^3∫x●dx/(1-x)^3∫x●dx/(1-x)^3

∫(2-3x)³dx ∫1/(2x+5)∧11dx

∫(2-3x)³dx∫1/(2x+5)∧11dx∫(2-3x)³dx∫1/(2x+5)∧11dx∫(2-3x)³dx∫1/(2x+5)∧11dx这么简单都不会么这么简单你

∫(x^3-x^2+x+1)/(x^2+1) dx∫(x+4)/(x^2-x-2) dx

∫(x^3-x^2+x+1)/(x^2+1)dx∫(x+4)/(x^2-x-2)dx∫(x^3-x^2+x+1)/(x^2+1)dx∫(x+4)/(x^2-x-2)dx∫(x^3-x^2+x+1)/(