<s>[INST] Clarification on partial fraction expansion I would like to use the cover up method for the following equation.
$$\frac{1}{x^2(x+1.79)}$$
and it breaks down into
$$\frac{A}{x}\quad\frac{B}{x^2}\quad\frac{C}{(x + 1.79)}$$
I realize that you cover up $(x+1.79)$ to get $x = -1.79$ but what do you do for the other variables?
Thanks for your help.
 [/INST] Here it is another way to approach for the sake of curiosity.
On one hand, we have that
\begin{align*}
\frac{1}{x^{2}(x+k)} & = \frac{1}{k}\left(\frac{k}{x^{2}(x+k)}\right)\\\\
& = \frac{1}{k}\left(\frac{(x + k) - x}{x^{2}(x+k)}\right)\\\\
& = \frac{1}{kx^{2}} - \frac{1}{kx(x+k)}
\end{align*}
On the other hand, we have that
\begin{align*}
\frac{1}{x(x+k)} & = \frac{1}{k}\left(\frac{k}{x(x+k)}\right)\\\\
& = \frac{1}{k}\left(\frac{(x + k) - x}{x(x+k)}\right)\\\\
& = \frac{1}{kx} - \frac{1}{k(x+k)}
\end{align*}
Gathering both results, it results that
\begin{align*}
\frac{1}{x^{2}(x+k)} = \frac{1}{kx^{2}} - \frac{1}{k^{2}x} + \frac{1}{k^{2}(x+k)}
\end{align*}
At your case, $k = 1.79$.
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