The Binomial Theorem for negative powers says that for $|x| < 1$
$$(1+x)^{-1} = 1 - x + x^2 + \mathcal{o}(x^2)$$
Therefore we have:
$$\frac 2{(2x-3)(2x+1)} = \frac 1{2(2x-3)} - \frac 1{2(2x+1)} = -\frac 16\left(1-\frac 23x\right)^{-1} - \frac 12\left(1+2x \right )^{-1} = -\frac16\left(1 + \frac 23x + \frac 49x^2\right)-\frac 12\left(1 - 2x + 4x^2 \right ) = \boxed{-\frac 23 + \frac 89 x - \frac{56}{27}x^2}$$
This holds for $|x| < \frac12$.
This is probably the wrong proof for you, but I will post it anyways. (requires calculus)
Note that $f(x)=(a+x)^n$ is an analytic function in $x$ for arbitrary $a,n$ since on its own, it is a power series with one term.
If it is an analytic function, then it should follow Taylor's theorem.
Now, if we take the expansion around $x=0$, we get
$$(a+x)^n=a^n+na^{n-1}x+\frac{n(n+1)}2a^{n-2}x^2+\dots$$
Since $f(0)=a^n$, $f'(0)=na^{n-1}$, $\dots f^{(k)}(0)=n(n+1)(n+2)\dots(n+k-1)a^{n-k}$
or
$$(a+x)^n=\sum_{k=0}^\infty\frac{n(n+1)(n+2)\dots(n+k-1)}{k!}a^{n-k}x^n$$
$$(a+x)^n=\sum_{k=0}^\infty\binom nka^{n-k}x^n$$
where $f'(x)$ is the first derivative of $f(x)$, $f''(x)$ the second derivative, etc. $f^{(k)}(x)$ is the $k$th derivative of $f(x)$.
Best Answer
For indices which are not positive integers you look at $(1+x)^a$ for $|x| \lt 1$ and expand as a power series in $x$. When $a$ is a positive integer the coefficient of $x^k$ is $\binom{a}{k}$. This may be written as:
$$ P_k(a) = \frac1{k!}a(a-1)...(a-k+1) $$ so that (still with $a$ a positive integer) we have the binomial expansion ($P_0(a)=1$): $$ (1+x)^a = \sum_{k=0}^a P_k(a)x^k $$ since $P_k(a) = 0$ if $k \gt a$ we may write this as: $$ (1+x)^a = \sum_{k=0}^{\infty} P_k(a)x^k $$ and it turns out that this same form can be used for fractional or negative integer values of $a$ for which $P_k(a) \ne 0$ for an infinite sequence of values of $k$.
To see why this should work let us compute: $$ (1+x)^{a+1} = (1+x)(1+x)^a $$ if the expansion is valid we require: $$ \sum_{k=0}^{\infty} P_k(a+1)x^k = (1+x)\sum_{k=0}^{\infty} P_k(a)x^k $$ or, for $k \gt 0$ $$ P_k(a+1) = P_k(a) + P_{k-1}(a)\tag{1} $$ In other words (leaving questions of convergence aside) we want the polynomials $P_k(a)$ to satisfy the same recurrence relation as the binomial coefficients do for $a$ an integer: $$ \binom{a+1}{k}=\binom{a}{k}+\binom{a}{k-1} $$ you may find it instructive to prove (1) directly. Or you could note that for each value of $k$ the relation is a polynomial equation in $a$ of degree $k$, which we already know is satisfied for an infinite set of (positive integer $\gt k$) values of $a$. so it must hold identically.