<s>[INST] 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.
 [/INST] We want to perform partial fraction of $\frac{6x}{x^3-8},$ therefore
$$\frac{6x}{x^3-8} = \frac{6x}{(x-2)(x^2+2x+4)} = \frac{Ax+B}{x^2+2x+4} + \frac{C}{x-2},$$
hence
$$\frac{6x}{(x-2)(x^2+2x+4)} = \frac{Ax^2-2Ax+Bx-2B+Cx^2+2Cx+4C}{(x-2)(x^2+2x+4)} = \frac{(A+C)x^2+(-2A+B+2C)x+(4C-2B)}{(x-2)(x^2+2x+4)},$$
and here is our system
$$
\begin{cases}
A+C = 0\\
2C+B-2A = 6\\
4C-2B=0
\end{cases},
$$
solving the system we obtain
$$
\begin{cases}
C = 1\\
B = 2\\
A=-1
\end{cases},
$$
can you do the integration, having $\frac{6x}{x^3-8} = \frac{2-x}{x^2+2x+4} + \frac{1}{x-2}?$
</s>