Integral using partial fractions I am having trouble using partial fractions to evaluate 
$$\int \frac{6x}{x^3-8} dx.$$  I can find the denominator but using equations to find the numerator is difficult.


Partial fractions for: $\frac{6x}{x^3-8}$

$$\frac{6x}{(x-2)(x^2+2x+4)}=\frac{C+Bx}{x^2+2x+4}+\frac{A}{x-2}$$
$$6x=A(x^2+2x+4)+(x-2)(C+Bx)$$
$$6x=4A-2C+(A+B)x^2+(2A-2B+C)x$$
$$\begin{cases}
4A-2C=0\\
2A-2B+C=6\\
A+B=0
\end{cases}$$
Solve

$$\int \frac{6x}{x^3-8}dx$$

$$=6\int\frac{dx}{(x-2)(x^2+2x+4)}\overset{\text{partial fractios}}{=}\int\frac{2-x}{x^2+2x+4}dx+\int\frac{dx}{x-2}$$
$$=-\frac 1 2\int\frac{2x+2}{x^2+2x+4}dx+3\int\frac{dx}{x^2+2x+4}+\int\frac{dx}{x-2}$$
Set $t=x^2+2x+4$ and $dt=(2x+2)dx$
$$=-\frac 1 2 \int \frac 1 t+3\int\frac{dx}{x^2+2x+4}+\ln|x-2|$$
$$\overset{\text{complete square}}{=}-\frac{\ln|t|}{2}+\ln|x-2|+3\int\frac{dx}{(x+1)^2+3}$$
$$=\color{red}{-\frac{\ln|x^2+2x+4|}{2}+\ln|x-2|+\sqrt 3\arctan\left(\frac{x+1}{\sqrt 3}\right)+\mathcal C}$$
