How to integrate $\int{\frac{6x}{x^3+8}dx}$ I'm having some trouble solving this integral using partial fraction method: $$\int{\frac{6x}{x^3+8}dx}.$$
After expanding $x^3+8$ into $(x-2)(x^2+2x+4)$ and expanding the original integral into $$\int{\frac{A}{x-2}+\frac{Bx+C}{x^2+2x+4}}dx,$$ I got $1$,$-1$, and $2$ for $A$, $B$, and $C$, respectively, yielding $$\int{\frac{1}{x-2}+\frac{-x+2}{x^2+2x+4}}dx.$$
This simplifies to $$\ln{|x-2|}-\int{\frac{x-2}{(x+1)^2+3}dx}.$$
How do I solve this resultant integral? That's where I'm stuck. 

Write the numerator of the last integral as $(x+1)-3,$ and break the integral up into the two parts. The first integral will be a logarithm the second an arctan but do the calculations yourself.
