The general problem of finding $\int \frac{dx}{(ax^2 + bx + c)^2}$ I know that $\displaystyle\int \dfrac1{ax^2+bx+c}\,\mathrm dx$ is easily solvable using completing the square, but my question is how would would find
$$\displaystyle\int\frac{1}{\left(ax^2+bx+c\right)^2}\,\mathrm dx$$
I have tried using the same approach as before but its not getting me anywhere... any help would be great!

If the quadratic factor ($a \neq 0$) that appears in the denominator can be factored into linear factors over the reals, then a partial fraction decomposition can be used. I will assume this is doable for you so will only concentrate on the case where the quadratic factor in the denominator is irreducible over the reals. In this case $a \neq 0$ and $b^2 < 4ac$. 
As you noted, one begins by completing the square. Doing so yields
$$I = \int \frac{dx}{(a x^2 + bx + c)^2} = \frac{1}{a^2} \int \frac{dx}{\left [\left (x + \frac{b}{2a} \right )^2 +  \left (\frac{c}{a} - \frac{b^2}{4a^2} \right )\right ]^2}.$$
Now let 
$$k^2 = \frac{c}{a} - \frac{b^2}{4a^2} > 0,$$
and is the case since $b^2 < 4ac$. Using a substitution of $u = x + b/(2a)$ we have
\begin{align*}
I &= \frac{1}{a^2} \int \frac{du}{(u^2 + k^2)^2}\\
&= \frac{1}{a^2 k^2} \int \frac{(u^2 + k^2) - u^2}{(u^2 + k^2)^2} \, du\\
&= \frac{1}{a^2 k^2} \int \frac{du}{u^2 + k^2} - \frac{1}{a^2 k^2} \int \frac{u^2}{(u^2 + k^2)^2}\\
&= \frac{1}{a^2 k^2} \left (I_1 - I_2 \right ).
\end{align*}
The first of these integrals can be readily found. The result is
$$I_1 = \frac{1}{k} \tan^{-1} \left (\frac{u}{k} \right ) + C_1 = \frac{1}{k} \tan^{-1} \left (\frac{2ax + b}{2ak} \right ) + C_1.$$
For the second integral, if we note that
$$\int \frac{2u}{(u^2 + k^2)^2} \, du = -\frac{1}{u^2 + k^2} + C,$$
integrating by parts we have
\begin{align*}
I_2 &= \int \frac{u^2}{(u^2 + k^2)^2} \, du\\
&= \frac{1}{2} \int \frac{2u}{(u^2 + k^2)^2} \cdot u \, du\\
&= -\frac{u}{2(u^2 + k^2)} + \frac{1}{2} \int \frac{du}{u^2 + k^2}\\
&= -\frac{u}{2(u^2 + k^2)} + \frac{1}{2k} \tan^{-1} \left (\frac{u}{k} \right ) + C_2\\
&= -\frac{2ax + b}{4(ax^2 + bx + c)} + \frac{1}{2k} \tan^{-1} \left (\frac{2ax + b}{2ak} \right ) + C_2.
\end{align*}
Thus
$$\boxed{\int \frac{dx}{(ax^2 + bx + c)^2} = \frac{1}{2a^2 k^3} \tan^{-1} \left (\frac{2ax + b}{2ak} \right ) + \frac{2ax + b}{4a^2 k^2(ax^2 + bx + c)} + C}$$
for the irreducible case where $a \neq 0$, $b^2 < 4ac$ such that $k = \sqrt{\dfrac{c}{a} - \dfrac{b^2}{4a^2}}$.
Comment
The above procedure can be readily generalised to the case
$$\int \frac{dx}{(ax^2 + bx + c)^n},$$
where $n \in \mathbb{N}$ and the factor appearing in the denominator is irreducible over the reals. 
