Partial fraction expansion problem

Question

I have a fraction written like so:

$$ \frac{3kX}{L\rho(3AX + kL)}. \qquad\qquad\qquad (1)$$

I have been told that a 'partial fraction expansion' can be used to re-write $(1)$ as

$$ \frac{k}{AL\rho} - \frac{k^2}{A\rho(3AX + kL)}. $$

However, I am really unsure about how this is achieved? What steps are required?

(The only variable in this expression is $k$. The rest are constants.)

Answer 1

We can first simplify the constants by the following

$$f(k)=\frac{3Xk}{L\rho(3AX+Lk)}=\frac{3Xk}{L^2\rho\left(\frac{3AX}{L}+k\right)}$$

And denote

$$\xi=\frac{3AX}{L}$$

So then

$$f(k)=\left(\frac{3X}{L^2\rho}\right)\frac{k}{\xi+k}$$

Where

$$\frac{3X}{L^2\rho}=\frac{\xi}{AL\rho}$$

There are many ways to rewrite $f(k)$, and to isolate one linear term in $k$ it can be rewritten in the form for some yet to be determined function $g(k)$

$$\frac{k}{\xi+k}=\frac{k}{\xi}+g(k)$$

Solve for $g(k)$ and combine the fractions

$$g(k)=\frac{k}{\xi+k}-\frac{k}{\xi}=-\frac{k^2}{\xi(\xi+k)}$$

Here the reason for dividing $k$ by $\xi$ is evident. Doing so simplifies the expression when the coefficient $\frac{\xi}{AL\rho}$ is reintroduced because solving for $g(k)$ yielded a fraction that is also divided by $\xi$ and in the numerator of $g(k)$, $\xi k-\xi k=0$, so there are no linear terms in $g(k)$

$$\therefore f(k)=\left(\frac{\xi}{AL\rho}\right)\left(\frac{k}{\xi}-\frac{k^2}{\xi(\xi+k)}\right)=\frac{k}{AL\rho}-\frac{k^2}{AL\rho\left(\frac{3AX}{L}+k\right)}$$

Answer 2

We have:

$$ \frac{3kX}{L\rho(3AX + kL)} = \frac{C_1k}{AL\rho} + \frac{C_2k^2}{A\rho(3AX + kL)} $$

$$\therefore 3kXA = C_1k(3AX+kL) + C_2k^2L. $$

For $k =\frac{-3AX}{L} $ we have: $$C_2 = -1$$ Using k=1:

$$ 3XA = C_1(3Ax+L) - L $$ $$\therefore C_1 = 1 $$

Substituting this you have:

$$ \frac{3kX}{L\rho(3AX + kL)} = \frac{k}{AL\rho} - \frac{k^2}{A\rho(3AX + kL)} $$