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?

After square completion and substituting $u=\frac{x-2}{3}$, there is a simple standard trick to evaluate the integral without trigonometric substitutions:
$$\int \dfrac{dx}{(x^2-4x+13)^2} \stackrel{u=\frac{x-2}{3}}{=}\frac 1{27} \underbrace{\int \frac{1}{(u^2+1)^2}du}_{I(u)}$$
Just rewrite the numerator
$$I(u) = \int\frac{1+u^2-u^2}{(u^2+1)^2}du = \arctan u - \frac 12\underbrace{\int u \frac{2u}{(u^2+1)^2}}_{J(u)}$$
So, only one quick partial integration gives
$$J(u) = -\frac u{u^2+1}+\arctan u$$
Hence,
$$I(u) = \arctan u - \frac 12\left(-\frac u{u^2+1}+\arctan u\right) =\frac 12 \left(\arctan u + \frac u{u^2+1}\right)$$
Finally, substitute back $u=\frac{x-2}{3}$ and you are done:
$$\int \dfrac{dx}{(x^2-4x+13)^2} = \frac 1{27}I(u)= \frac 1{54}\left(\arctan \frac{x-2}{3} + \frac{\frac{x-2}{3}}{\left(\frac{x-2}{3}\right)^2+1}\right) (+C)$$
$$= \frac 1{54}\left(\arctan \frac{x-2}{3} + \frac{3(x-2)}{\left(x-2\right)^2+9}\right)(+C)$$
