Write the function as a sum of partial fractions with parametric numerators, using as denominators all the factors of the denominator of the rational function:
1/(x^3(x-4)) = A/x^3+B/x^2+C/x+D/(x-4)1x3(x−4)=Ax3+Bx2+Cx+Dx−4
Execute the sum:
1/(x^3(x-4)) = (A(x-4) +Bx(x-4)+Cx^2(x-4) +Dx^3)/(x^3(x-4))1x3(x−4)=A(x−4)+Bx(x−4)+Cx2(x−4)+Dx3x3(x−4)
Since the denominators are equal we can equate the numerators:
Ax -4A + Bx^2-4Bx +Cx^3-4Cx^2 +Dx^3 =1 Ax−4A+Bx2−4Bx+Cx3−4Cx2+Dx3=1
(C+D)x^3 + (B-4C)x^2 +(A-4B)x -4A = 1(C+D)x3+(B−4C)x2+(A−4B)x−4A=1
Two polynomials are equal if the coefficients of the same degree in xx are equal, so:
{(-4A =1),(A-4B=0),(B-4C=0), (C+D=0):}
{(A =-1/4),(B=A/4=-1/16),(C=B/4=-1/64), (D=-C=1/64):}
Then:
1/(x^3(x-4)) = -1/(4x^3)-1/(16x^2)-1/(64x)+1/(64(x-4))
and integrating:
int dx/(x^3(x-4)) = -int dx/(4x^3)-int dx/(16x^2)-int dx /(64x)+int dx/(64(x-4))
int dx/(x^3(x-4)) =1/(8x^2)+1/(16x)-64lnabs x+64ln abs(x-4)+C
int dx/(x^3(x-4)) =(x+2)/(16x^2)+64lnabs((x-4)/x)+C
