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)#
Execute the sum:
#1/(x^3(x-4)) = (A(x-4) +Bx(x-4)+Cx^2(x-4) +Dx^3)/(x^3(x-4))#
Since the denominators are equal we can equate the numerators:
#Ax -4A + Bx^2-4Bx +Cx^3-4Cx^2 +Dx^3 =1 #
#(C+D)x^3 + (B-4C)x^2 +(A-4B)x -4A = 1#
Two polynomials are equal if the coefficients of the same degree in #x# 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#
