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?

Let us ise Euler's transformation
$$I=\int \frac{dx}{(x^2-4x+13)^2}=\int \frac{dx}{(x-a)^2 (x-b)^2},~~a,b=2\pm 3i.$$
Let $$t=\frac{x-a}{x-b} \implies x=\frac{bt-a}{t-1} \implies dx=\frac{a-b}{(t-1)^2}.$$
Then $$I=(b-a)^{-3} \int \frac{dt}{t^2(t-1)^2}=(a-b)^{-3}\int \frac{u^2 du}{(u-1)^2}, u=1/t .$$
Next use $u=v+1$, then
$$I=(a-b)^3 \int [1-2/v+1/v^2] dv= (a-b)^{-3}[v-2 \ln v -1/v] $$
$$I=(a-b)^{-3}\left(\frac{a-b}{x-a}-2\ln \frac{a-b}{x-a}-\frac{x-a}{a-b}\right)$$
