How do you use partial fractions to find the integral #int (6x)/(x^3-8)dx#?

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)=color(green)(intdx/(x-2))-int((x-2)dx)/(x^2+2x+4)#

We integrate separately

#color(green)(intdx/(x-2))=color(green)(ln(∣x-2∣))#

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

#=color(red)(int((x+2)dx)/(x^2+2x+4))-color(blue)(int(4dx)/(x^2+2x+4))#

Let #u=x^2+2x+4#, #=>#, #du=(2x+2)dx#

Therefore,

#color(red)(int((x+2)dx)/(x^2+2x+4)=1/2int(du)/u=1/2lnu)#

#color(red)(=1/2ln(∣x^2+2x+4∣))#

#x^2+2x+4=x^2+2x+1+3=(x+1)^2+3#

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

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

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

Let #tantheta=(x+1)/sqrt3##=># #sec^2theta d theta=dx/sqrt3#

Therefore,

#color(blue)(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#

#color(blue)(4/3int(dx)/((((x+1)/sqrt3)^2+1))=4/sqrt3intd theta)#

#color(blue)(=4/sqrt3arctan((x+1)/sqrt3))#