Factors of #1-x^3# are #(1-x)(1+x+x^2)#, hence partial fractions of #(2x)/(1-x^3)# are of the form
#(2x)/(1-x^3)=A/(1-x)+(Bx+C)/(1+x+x^2)#
i.e. #(2x)/(1-x^3)=(A(1+x+x^2)+(Bx+C)(1-x))/(1-x^3)#
Comparing coefficients of constant term, #x# and #x^2# in numerator
#A+C=0#, #A+B-C=2# and #A-B=0#
solving these we get #A=B=2/3#, #C=-2/3# and hence
#(2x)/(1-x^3)=2/(3(1-x))+(2x-2)/(3(1+x+x^2))# and
#int(2x)/(1-x^3)dx=int2/(3(1-x))dx+int(2x-2)/(3(1+x+x^2))dx#
Now #int2/(3(1-x))dx=2/3int1/(1-x)dx=-2/3ln|x-1|#
and #int(2x-2)/(3(1+x+x^2))dx#
= #int(2x+1)/(3(1+x+x^2))dx-int3/(1+x+x^2)dx#
= #1/3ln|1+x+x^2|-3[2/sqrt3tan^(-1)((2x+1)/sqrt3)]#
For former observe #d/(dx)(1+x+x^2)=2x+1# and for latter, we can have
#int3/(1+x+x^2)dx=3int1/((x+1/2)^2+3/4)dx=3int1/((x+1/2)^2+(sqrt3/2)^2)dx#
= #3[2/sqrt3tan^(-1)((2x+1)/sqrt3)]#
Hence #int(2x)/(1-x^3)dx=-2/3ln|x-1|+1/3ln|1+x+x^2|-3[2/sqrt3tan^(-1)((2x+1)/sqrt3)]#
