Clarification on partial fraction expansion

Question

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.

Answer 1

You can also use cover up method to compute $B$.

To compute $A$, note that

$$1=Ax(x+1.79)+B(x+1.79)+Cx^2$$

You can let $x$ take some other value to compute $A$ or you can just compare the coefficient.

We have $A+C=1$, since we already know $C$, we have $A=1-C$.

Answer 2

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, $k = 1.79$.

Answer 3

With $$\frac{1}{x^2(x+1.79)}= \frac{A}{x}+\frac{B}{x^2}+\frac{C}{x + 1.79}$$ the coefficients are calculated as follows

\begin{align} & C =\lim_{x\to -1.79} \frac{x+1.79}{x^2(x+1.79)}=\frac1{1.79^2}\\ & B =\lim_{x\to 0} \frac{x^2}{x^2(x+1.79)}=-\frac1{1.79}\\ &A = \frac{x}{x^2(x+1.79)}-\frac {Bx}{x^2}- \frac {Cx}{x+1.79}=\frac1{1.79^2} \end{align}