How do you integrate #int (2x)/(1-x^3)# using partial fractions?

As #1-x^3=(1-x)(1+x+x^2)# let

#(2x)/(1-x^3)=A/(1-x)+(Bx+C)/(1+x+x^2)#

= #(A+Ax+Ax^2-Bx^2+Bx-Cx+C)/(1-x^3)#

= #((A-B)x^2+(A+B-C)x+(A+C))/(1-x^3)#

Hence #A-B=0# i.e. #B=A#, #A+C=0# i.e. #C=-A#

and #A+B-C=2# i.e. #A+A+A=2# i.e. #A=2/3#

and #B=2/3# and #C=-2/3#

and #int(2x)/(1-x^3)dx=-2/3int(dx)/(x-1)+2/3int(x-1)/(x^2+x+1)#

= #-2/3int(dx)/(x-1)+1/3int(2x+1)/(x^2+x+1)dx+int1/(x^2+x+1)dx#

= #-2/3ln|x-1|+1/3ln|x^2+x+1|+int1/(x^2+x+1)dx#

Now #int1/(x^2+x+1)dx=int1/((x+1/2)^2+3/4)dx#

and putting #u=x+1/2# and #du=dx# above becomes

#int1/(u^2+3/4)du# and as #int1/(x^2+a^2)dx=1/atan^(-1)(x/a)#

Hence #int1/(u^2+3/4)du=1/(sqrt3/2)tan^(-1)(u/(sqrt3/2))#

and #int1/(x^2+x+1)dx=2/sqrt3tan^(-1)((2x+1)/sqrt3)#

and #int(2x)/(1-x^3)dx=-2/3ln|x-1|+1/3ln|x^2+x+1|+2/sqrt3tan^(-1)((2x+1)/sqrt3)+c#

Factorize the denominator:

#(1-x^3) = (1-x)(1+x+x^2)#

Write the integrand as:

#x/(1-x^3) = A/(1-x)+ (Bx+C)/(1+x+x^2)#

#x/(1-x^3) = (A(1+x+x^2)+(Bx+C)(1-x))/((1-x)(1+x+x^2))#

As the denominators are equal, so must be the numerators:

#x = A+Ax+Ax^2+Bx+C-Bx^2-Cx#

#x = (A-B)x^2 + (A+B-C)x+(A+C)#

and equating the coefficients of the same degree in #x#:

#{(A-B=0),(A+B-C=1),(A+C=0):}#

#{(A=B),(2A-C=1),(C=-A):}#

#{(A=1/3),(B=1/3),(C=-1/3):}#

Then:

#x/(1-x^3) = 1/3 1/(1-x)+ 1/3 (x-1)/(1+x+x^2)#

and:

#int (2x)/(1-x^3)dx = 2/3int dx/(1-x) +2/3int (x-1)/(1+x+x^2)dx#

Solve separately the integrals:

#2/3int dx/(1-x) = - 2/3int (d(1-x))/(1-x) = - 2/3ln abs (1-x) + C_1#

Split the other noting that #d(1+x+x^2) = 1+2x#:

#2/3int (x-1)/(1+x+x^2)dx = 1/3 int (2x+1-3)/(1+x+x^2)dx#

#2/3int (x-1)/(1+x+x^2)dx = 1/3int (2x+1)/(1+x+x^2)dx - int 1/(1+x+x^2)dx #

Now solve:

#1/3int (2x+1)/(1+x+x^2)dx = 1/3int (d(1+x+x^2))/(1+x+x^2) = 1/3ln abs (1+x+x^2) + C_2#

and finally:

#int 1/(1+x+x^2)dx = int dx/(3/4+ (x+1/2)^2)#

#int 1/(1+x+x^2)dx = 4/3 int dx/(1+ ((2x+1)/sqrt3)^2)#

#int 1/(1+x+x^2)dx = 2/sqrt3 int (d((2x+1)/sqrt3))/(1+ ((2x+1)/sqrt3)^2)#

#int 1/(1+x+x^2)dx = 2/sqrt3 arctan((2x+1)/sqrt3)+C_3#

Putting it all together:

#int (2x)/(1-x^3)dx = -2/3 ln abs (1-x) +1/3 ln abs (1+x+x^2) -2/sqrt3arctan((2x+1)/sqrt3)+C#