How do you integrate #(6x)/(x^3-8)# using partial fractions?

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)#