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((1%x^2)^3/2)x^%8不定积分

这就是一个很简单的三角换元,令x=sint,则dx=costdt,∫(1-x^2)^(3/2) dx=∫cost(1-(sint)^2)^3/2dt=∫(cost)^4dt=∫((cos4t)/8+(cos2t)/2+3/8)dt(二倍角公式得到的)=-(sin4t)/32-(sin2t)/4+3t/8=-sintcost(1-2(sint)^2)/8-sintcost/2+3t/3(还是二倍角)=-x(1-x^2)^1/2(5-2x^2)/8+3arcsinx/8

1/(x^2+x^3)=x^(-2)+x^(-3) x^(-2)的不定积分=-x^(-1); x^(-3)的不定积分=-1/2*x^(-2).1/(x^2+x^3)不定积分是:-x^(-1)-1/2*x^(-2)+常数C.

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^2/3+x=1/3(x^2+3x)=1/3[(x+3/2)^2-9/4]=1/3*9/4[4/9(x+3/2)^2-1]=3/4[(2x/3+1)^2-1]则:1/(x^2/3+x)=(4/3)/[(2x/3+1)^2-1]设2x/3+1=sect t=arcsec(2x/3+1)2/3dx=tan^2tdt dx=3/2tan^2tdt代入原式:4/3 *3/

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

原式=∫ (x-3x+3x-1)/x dx=∫ (x-3+3/x-1/x)dx=x/2-3x+3ln|x|-1/x+C

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^2/1+3x^2 dx=∫[(1/3+x^2)-1/3]/(1+3x^2) dx=∫1/3-[1/3/(1+3x^2)] dx=1/3x-1/3∫1/(1+3x^2) dx=1/3x-1/(3√3)∫1/(1+3x^2) d√3x=1/3x-1/(3√3)arctan(√3x)+c

该题属于有理函数的积分,有一整套适用的解法(不过你这一题的数字太不人性化,改一下可省敲几个字),给你过程: 1)先用待定系数法将有理函数 1/(x^3+8)^2 =1/{(x+2)[(x^2)-2x+4]}^2分解为最简分式之和: 1/(x^3+8)^2 = A/(x+2) + B/(x+2)^2 + (Cx+D)/[(x^2)-2x+4] + (Ex+F)/[(x^2)-2x+4]^2; 2)再行积分……

∫x^2 * arctanxdx =1/3*∫arctanxd(x^3) =1/3*x^3*arctanx-1/3*∫x^3/(1+x^2) =1/3*x^3*arctanx-1/3*∫(x^3+x-x)/(1+x^2) =1/3*x^3*arctanx-1/3*∫xdx+1/3*∫x/(1+x^2) =1/3*x^3*arctanx-1/6*x^2+1/6*ln(1+x^2)+c

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