partial FRACTION!!!!!

Long time problem solver, first time solution provider. Let me know if I can clarify.

Step 1 : Typically partial fraction decomposition begins with combining the two fractions into one.

$\frac{\alpha}{2x+3} + \frac{\beta}{2x-1}$ = $\frac{\alpha(2x-1)}{(2x+3)(2x-1)} + \frac{\beta(2x+3)}{(2x+3)(2x-1)}$

$= \frac{\alpha(2x-1) + \beta(2x+3)}{(2x+3)(2x-1)}$

Step 2 : Now, to the extent possible, we want to make this new fraction look as similar to the one we started with is as possible. This will help us to determine what $\alpha$ and $\beta$ should be.

$\frac{\alpha(2x-1) + \beta(2x+3)}{(2x+3)(2x-1)} = \frac{\alpha \cdot 2x - \alpha + \beta \cdot 2x + 3\beta}{(2x+3)(2x-1)}$ )

$= \frac{\alpha \cdot 2x + \beta \cdot 2x - \alpha + 3\beta}{(2x+3)(2x-1)}$ )

$= \frac{2x(\alpha + \beta) - \alpha + 3\beta}{(2x+3)(2x-1)}$ )

We now have something pretty close where the $x$ has $\alpha$ and $\beta$ constants.

Step 3 : We can now start to make some comparisons.

Let's think of $2x(\alpha + \beta)$ as equal to 6x) and see if we can solve for just \(\alpha and $\beta$

$2x(\alpha + \beta) = 6x$

divide both sides by $2x$ and we have : $\alpha + \beta = 3$

That's what we want! 3. We could keep going to solve for $\alpha$ and $\beta$ , but we have our answer.