The denominator is
#x^3-8=(x-2)(x^2+2x+4)#
Therefore,
#(6x)/(x^3-8)=(6x)/((x-2)(x^2+2x+4))#
#=A/(x-2)+(Bx+C)/(x^2+2x+4)#
#=(A(x^2+2x+4)+(Bx+C)(x-2))/((x-2)(x^2+2x+4))#
So,
#6x=A(x^2+2x+4)+(Bx+C)(x-2)#
Let #x=2#, #=>#, #12=12A#, #=>#, #A=1#
Let #x=0#, #=>#, #0=4A-2C#, #=>#, #C=2#
Coefficients of #x^2#, #=>#, #0=A+B#, #=>#, #B=-1#
Therefore,
#(6x)/(x^3-8)=1/(x-2)+(-x+2)/(x^2+2x+4)#
So,
#int(6xdx)/(x^3-8)=intdx/(x-2)-int((x-2)dx)/(x^2+2x+4)#
We integrate separately
#intdx/(x-2)=ln(∣x-2∣)#
#int((x-2)dx)/(x^2+2x+4)=int((x+2-4)dx)/(x^2+2x+4)#
#=int((x+2)dx)/(x^2+2x+4)-int(4dx)/(x^2+2x+4)#
Let #u=x^2+2x+4#, #=>#, #du=(2x+2)dx#
Therefore,
#int((x+2)dx)/(x^2+2x+4)=1/2int(du)/u=1/2lnu#
#=1/2ln(∣x^2+2x+4∣)#
#x^2+2x+4=x^2+2x+1+3=(x+1)^2+3#
#int(4dx)/(x^2+2x+4)=4int(dx)/((x+1)^2+3)#
#=4int(dx)/(3(((x+1)/sqrt3)^2+1))#
#=4/3int(dx)/((((x+1)/sqrt3)^2+1))#
Let #tantheta=(x+1)/sqrt3##=># #sec^2theta d theta=dx/sqrt3#
Therefore,
#4/3int(dx)/((((x+1)/sqrt3)^2+1))=4/3int(sqrt3 sec^2 theta d theta)/(1+ tan^2theta)#
But #1+tan^2 theta=sec^2theta#
#4/3int(dx)/((((x+1)/sqrt3)^2+1))=4/sqrt3intd theta#
#=4/sqrt3arctan((x+1)/sqrt3)#