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.

First, factor $x^3-8$. $x^3-8=(x^2+2x+4)(x-2)$
$$\frac{6x}{x^3-8}=\frac{ax+b}{x^2+2x+4}+\frac{c}{x-2}$$
Then multiply the lhs denominator out on the right side:
$$6x=(x-2)(ax+b)+c(x^2+2x+4)$$
When $x=2$, $(x-2)(ax+b)$ vanishes and $$6(2)=c(4+4+4)\Rightarrow c=1$$
Now, we can use $c$ to solve for $ax+b$.
$$6x-(1)(x^2+2x+4)=(x-2)(ax+b)$$
which gives us $a=-1$ and $b=2$.
Thus $$\int\frac{6x}{x^3-8}dx=\int\frac{2-x}{x^2+2x+4}+\frac{1}{x-2}dx$$
Which gives you what echzhen just demonstrated.
