Here's what I've done:
$\frac{x+1}{2x-3}<\frac{1}{x-3}$
$x+1<\frac{2x-3}{x-3}$
$(x+1)(x-3)<2x-3$
$x^2-2x-3<2x-3$
$x^2-4x<0$
$x(x-4)<0$
$0<x<4$
However this clearly fails because when $x$ is $2$, for example, the inequality fails.
Where have I gone wrong?
On
Beware of cross-multiplying when solving inequalities! You might be multiplying by a negative number – which would invalidate the inequality.
One way to proceed is to bring everything to one side and simplify:
$\quad\quad\quad\dfrac{x+1}{2x+3}\ <\ \dfrac1{x-3}$
$\implies\ 0\ <\ \dfrac1{x-3}-\dfrac{x+1}{2x+3}$
$\implies\ 0\ <\ \dfrac{2x+3-(x+1)(x-3)}{(x-3)(2x+3)}$
…etc.
Taking in account the comment of @SteamyRoot and resolve it again, you will have the answer $x\in I=]0,\frac{3}{2}[~\cup~]3,4[$.
Clearly the example you have taken ($x=2$) doesn't belong to I.