www.ypnh.net > 求x^3/(1+x^8)^2的不定积分,知道的告诉一下,要过程,谢谢了

求x^3/(1+x^8)^2的不定积分,知道的告诉一下,要过程,谢谢了

解:∫x*lnx/(1+x^2)^(3/2)dx=1/2*∫lnx*d(1+x^2)/(1+x^2)^(3/2)dx=1/2*∫lnx*d[-2/(1+x^2)^(1/2)]=-∫lnx*d[1/(1+x^2)^(1/2)]=∫[1/(1+x^2)^(1/2)]*1/x*dx-(lnx)/[(1+x^2)^(1/2)]积分式中令x=tant,则dx=sec^2tdt=∫1/sect*cott*sec^2tdt-(lnx)/[(1+x^2)^(1/2)]=∫csctdt-(lnx)

∫(x^3)/(1+x^8)^2 dx =(1/4)∫1/(1+x^8)^2d(x^4)【令t=x^4】 =(1/4)∫1/(1+t^2)^2dt【令t=tanu】 =(1/4)∫[1/(secu)^4][(secu)^2du] =(1/4)∫(cosu)^2du =(1/8)∫(1+cos2u)du =(1/8)[u+(sin2u)/2]+C =[arctan(x^4)]/8+x^4/[8(1+x^8)]+C.

解:∫√[(1-x)/(1+x)]dx=∫√[(1-x)/(1+x)(1-x)]dx=∫[(1-x)/√(1-x)]dx=∫[1/√(1-x)]dx -∫[x/√(1-x)]dx=arcsinx +√(1-x) +c

∫ x^3/(-2+x^8) dx令 u = x^4 则 du = 4 x^3 dx: = 1/4 ∫ 1/(u^2-2) du已知 ∫ of 1/(u^2-2) = -(tanh^(-1)(u/sqrt(2)))/sqrt(2):原式 = -(tanh^(-1)(u/sqrt(2)))/(4 sqrt(2))+C代回 u = x^4:原式 = -(tanh^(-1)(x^4/sqrt(2)))/(4 sqrt(2))+C等价于:(ln(sqrt(2)-x^4)-ln(x^4+sqrt(2)))/(8 sqrt(2))+C

由u=tant可解得:sint=u/√(1+u),cost=1/√(1+u)sin2t=2sintcost=2u/(1+u)=2x^4/(1+x^8)欢迎追问.如满意,请采纳.

x/(x+1)=[x(x+1)-x]/(x+1)=x-x/(x+1)=x-[x(x+1)-x]/(x+1)=x-x+x/(x+1)=x-x+(x+1-1)/(x+1)=x-x+1-1/(x+1)∴∫x/(x+1)dx=∫[x-x+1-1/(x+1)]dx=x/3-x/2+x-ln|x+1|+C

解:由题意可得: ∫x^3/(x+1)^8dx=∫(x^3+1-1)/(x+1)^8dx =∫[(x+1)(x^2-x+1)/(x+1)^8-1/(x+1)^8]dx =∫(x^2-x+1)/(x+1)^7dx-∫1/(x+1)^8dx =∫(x^2+2x+1-3x)/(x+1)^7dx-∫1/(x+1)^8dx =∫(x+1)^2/(x+1)^7dx-∫3x/(x+1)^7dx-∫1/(x+1)^8dx =∫1/(x+1)^5dx-∫(3x+3-3)/(x+

若是∫(1+x)/x^8 dx= ∫(x^-8 + x^-6) dx= x^(-8+1) / (-8+1) + x^(-6+1) / (-6+1) + c= -1/[7x^7] - 1/[5x^5]= -[7x+5] / [35x^7] + c若是∫dx/[x^8(1+x)]令1/[x^8*(1+x)] = a/x^8 + b/x^6 + c/x^4 + d/x^2 + e/(x+1)待定系数法,过程太烦,召唤答案~~a = 1,b = -

x^3=X(X^2+1)-X原式=[X(X^2+1)-X]/(x^2+1)∫[X(X^2+1)-X]/(x^2+1)dx=∫xdx-∫x/(x^2+1)dx=(1/2)x^2-(1/2)ln(1+x^2)+C不定积分=(1/2)x^2-(1/2)ln(1+x^2)代入负无穷到正无穷

1/(x^8-2)=1/[(x^4-√2)*(x^4+√2)]=√2/4*[1/(x^4-√2)-1/(x^4+√2)]∫x^3/(x^8-2) dx=∫1/(x^8-2) d(x^4/4)=1/4*∫1/(x^8-2) d(x^4)=1/4*∫√2/4*[1/(x^4-√2)-1/(x^4+√2)] d(x^4)=√2/16*∫[1/(x^4-√2)-1/(x^4+√2)] d(x^4)=√2/16*ln|x^4-√2|-√2/16*ln|x^4+√2| +C

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