#x^3+8=(x+2)(x^2-2x+4)#
Perform the decomposition into partial fractions
#1/(x^3+8)=1/((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))#
The denominators are the same, compare the numerators
#1=A(x^2-2x+4)+(Bx+C)(x+2)#
Let #x=-2#, #=>#, #1=12A#, #=>#, #A=1/12#
Let #x=0#, #=>#, #1=4A+2C#, #=>#, #C=1/3#
Coefficients of #x^2#
#0=A+B#, #=>#, #B=-1/12#
Therefore,
#1/(x^3+8)=(1/12)/((x+2))+(-1/12x+4/12)/(x^2-2x+4)#
#int(1dx)/(x^3+8)=int(1/12dx)/((x+2))+int((-1/12x+4/12)dx)/(x^2-2x+4)#
#=1/12ln(|x+2|)-1/12int((x-4)dx)/(x^2-2x+4)#
Let #x-4=1/2(2x-2)-3#
Perform the second integral separately
#int((x-4)dx)/(x^2-2x+4)=int(1/2(2x-2)dx)/(x^2-2x+4)-3int(dx)/(x^2-2x+4)#
#=1/2ln(|x^2-2x+4|)-3int(dx)/(x^2-2x+4)#
#x^2-2x+4=x^2-2x+1+3=(x-1)^2+(sqrt3)^2#
#=sqrt3(((x-1)/sqrt3)^2+1)#
So,
#3int(dx)/(x^2-2x+4)=3/sqrt3int(dx)/(((x-1)/sqrt3)^2+1)#
#=sqrt3arctan((x-1)/sqrt3)#
Finally,
#int(1dx)/(x^3+8)=1/12ln(|x+2|)-1/24ln(|x^2-2x+4|)+sqrt3/12arctan((x-1)/sqrt3)+C#
