<s>[INST] Integral: $\int \dfrac{dx}{(x^2-4x+13)^2}$? 
How can I integrate $$\int \dfrac{dx}{(x^2-4x+13)^2}?$$

Here is my attempt:
$$\int \dfrac{dx}{(x^2-4x+13)^2}=\int \dfrac{dx}{((x-2)^2+9)^2}$$
Substitute $x-2=3\tan\theta$, $\ dx=3\sec^2\theta d\theta$
\begin{align*}
&=\int \dfrac{3\sec^2\theta d\theta}{(9\tan^2\theta+9)^2}\\
&=\int \dfrac{3\sec^2\theta d\theta}{81\sec^4\theta}\\
&=\dfrac{1}{27}\int \cos^2\theta d\theta\\
&=\dfrac{1}{27}\int \frac{1+\cos2\theta}{2} d\theta\\
&=\dfrac{1}{54}\left(\theta+\frac{\sin2\theta}{2}\right)+C
\end{align*}
This is where I got stuck. How can I get the answer in terms of $x$?
Can I solve it by other methods?
 [/INST] Substitute $t=x-2$
\begin{align}
\int \dfrac{dx}{(x^2-4x+13)^2}
& =\int \dfrac{dt}{(t^2+9)^2}= \int \frac1{18t}d\left( \frac{t^2}{t^2+9}\right)\\
&=  \frac t{18(t^2+9)}+\frac1{18}\int \frac{dt}{t^2+9}\\
&= \frac t{18(t^2+9)}+\frac1{54}\tan^{-1}\frac t3+C
\end{align}
</s>