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.
