First, set-up:

$\displaystyle \displaystyle \frac{2}{x(x^2+5)} = \frac{a}{x}+\frac{bx+c}{x^2+5}$

You can see that if you cover-up $\displaystyle (x^2+5)$, then $\displaystyle a = \frac{2}{5}.$ But that is not going to give us b and c, so let's do it in the saver way.

Put the RHS over a common denomintor:

$\displaystyle \displaystyle \frac{2}{x(x^2+5)} = \frac{a(x^2+5)+(bx+c)(x)}{x(x^2+5)}

$

We have $\displaystyle 2 = a(x^2+5)+(bx+c)(x)$

[Stop for a moment. Let $\displaystyle x = 0$: Then we have $\displaystyle 2 = 5a$. So $\displaystyle a = \frac{2}{5}$. The cover-up method was really this in disguise. You can see why it doesn't work for finding b and c -- there is no value of x which can make $\displaystyle a(x^2+5)$ equal to zero. So if in general you have a partial fraction form of $\displaystyle \frac{Ax+B}{ax^2+bx+c}$, then you cannot find A and B by covering-up -- in other words, the denominator have to be linear for the cover-up to work.]

We have:

$\displaystyle 2 = ax^2+5a+bx^2+cx$ (expanding the right-hand side).

$\displaystyle \Rightarrow 2 = ax^2+bx^2+cx+5a$ (writing it in terms of descending powers of x).

$\displaystyle

\Rightarrow 2 = (a+b)x^2+(c)x+5a$ (factoring out the x's out).

Now, compare the coefficients:

We have $\displaystyle c = 0$.

$\displaystyle 5a = 2[/Math], so [Math]a = \frac{2}{5}$.

$\displaystyle a+b = 0$, and since $\displaystyle a = \frac{2}{5}$, we get $\displaystyle \frac{2}{5}+b = 0$, hence $\displaystyle b= -\frac{2}{5}$.

Therefore $\displaystyle \displaystyle \frac{2}{x(x^2+5)} = \frac{2}{5x}-\frac{2x}{5(x^2+5)}.$

It should of course take a much smaller space to write-up, but that's Mr Beefcake's attempt at being clear.