求x^4/（x^3+1）的不定积分要详细过程谢谢

求x^4/（x^3+1）的不定积分要详细过程谢谢
求x^4/（x^3+1）的不定积分
要详细过程谢谢

求x^4/（x^3+1）的不定积分要详细过程谢谢
看如下过程,有理分式会拆解吧?

Take the integral:
integral x^4/(x^3+1) dx
For the integrand x^4/(x^3+1), do long division:
 =   integral ((-x-1)/(3 (x^2-x+1))+x+1/(3 (x+1))) dx
Integrate the sum term by term and factor out constants:
 =  1/3 integral (-x-1)/(x^2-x+1) dx+ integral x dx+1/3 integral 1/(x+1) dx
Rewrite the integrand (-x-1)/(x^2-x+1) as -(2 x-1)/(2 (x^2-x+1))-3/(2 (x^2-x+1)):
 =  1/3 integral (-(2 x-1)/(2 (x^2-x+1))-3/(2 (x^2-x+1))) dx+ integral x dx+1/3 integral 1/(x+1) dx
Integrate the sum term by term and factor out constants:
 =  -1/2 integral 1/(x^2-x+1) dx-1/6 integral (2 x-1)/(x^2-x+1) dx+ integral x dx+1/3 integral 1/(x+1) dx
For the integrand (2 x-1)/(x^2-x+1), substitute u = x^2-x+1 and  du = (2 x-1)  dx:
 =  -1/6 integral 1/u du-1/2 integral 1/(x^2-x+1) dx+ integral x dx+1/3 integral 1/(x+1) dx
For the integrand 1/(x^2-x+1), complete the square:
 =  -1/6 integral 1/u du-1/2 integral 1/((x-1/2)^2+3/4) dx+ integral x dx+1/3 integral 1/(x+1) dx
For the integrand 1/((x-1/2)^2+3/4), substitute s = x-1/2 and  ds =   dx:
 =  -1/2 integral 1/(s^2+3/4) ds-1/6 integral 1/u du+ integral x dx+1/3 integral 1/(x+1) dx
Factor 3/4 from the denominator:
 =  -1/2 integral 4/(3 ((4 s^2)/3+1)) ds-1/6 integral 1/u du+ integral x dx+1/3 integral 1/(x+1) dx
Factor out constants:
 =  -2/3 integral 1/((4 s^2)/3+1) ds-1/6 integral 1/u du+ integral x dx+1/3 integral 1/(x+1) dx
For the integrand 1/((4 s^2)/3+1), substitute p = (2 s)/sqrt(3) and  dp = 2/sqrt(3)  ds:
 =  -1/sqrt(3) integral 1/(p^2+1) dp-1/6 integral 1/u du+ integral x dx+1/3 integral 1/(x+1) dx
The integral of 1/(p^2+1) is tan^(-1)(p):
 =  -(tan^(-1)(p))/sqrt(3)-1/6 integral 1/u du+ integral x dx+1/3 integral 1/(x+1) dx
The integral of 1/u is log(u):
 =  -(tan^(-1)(p))/sqrt(3)-(log(u))/6+ integral x dx+1/3 integral 1/(x+1) dx
For the integrand 1/(x+1), substitute w = x+1 and  dw =   dx:
 =  -(tan^(-1)(p))/sqrt(3)-(log(u))/6+1/3 integral 1/w dw+ integral x dx
The integral of 1/w is log(w):
 =  -(tan^(-1)(p))/sqrt(3)-(log(u))/6+(log(w))/3+ integral x dx
The integral of x is x^2/2:
 =  -(tan^(-1)(p))/sqrt(3)-(log(u))/6+(log(w))/3+x^2/2+constant
Substitute back for w = x+1:
 =  -(tan^(-1)(p))/sqrt(3)-(log(u))/6+x^2/2+1/3 log(x+1)+constant
Substitute back for p = (2 s)/sqrt(3):
 =  -(tan^(-1)((2 s)/sqrt(3)))/sqrt(3)-(log(u))/6+x^2/2+1/3 log(x+1)+constant
Substitute back for s = x-1/2:
 =  -(log(u))/6+x^2/2+1/3 log(x+1)-(tan^(-1)((2 x-1)/sqrt(3)))/sqrt(3)+constant
Substitute back for u = x^2-x+1:
 =  x^2/2-1/6 log(x^2-x+1)+1/3 log(x+1)-(tan^(-1)((2 x-1)/sqrt(3)))/sqrt(3)+constant
Which is equal to:
Answer: |  
|   =  1/6 (3 x^2-log(x^2-x+1)+2 log(x+1)-2 sqrt(3) tan^(-1)((2 x-1)/sqrt(3)))+constant

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