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?

You can solve it by using induction formula: $\color{blue}{\int \frac{dt}{(t^2+a^2)^n}=\frac{t}{2(n-1)a^2(t^2+a^2)^{n-1}}+\frac{2n-3}{2(n-1)a^2}\int \frac{dt}{(t^2+a^2)^{n-1}}}$   as follows
$$\int \dfrac{dx}{(x^2-4x+13)^2}$$$$=\int \dfrac{d(x-2)}{((x-2)^2+3^2)^2}$$
$$=\frac{x-2}{2\cdot 3^2((x-2)^2+3^2)}+\frac{1}{2\cdot 3^2}\int \frac{d(x-2)}{(x-2)^2+3^2}$$
$$=\frac{x-2}{18(x^2-4x+13)}+\frac{1}{18}\left(\frac13\tan^{-1}\left(\frac{x-2}{3}\right) \right)+C$$
$$=\bbox[15px,#ffd,border:1px solid green]{\frac{x-2}{18(x^2-4x+13)}+\frac{1}{54}\tan^{-1}\left(\frac{x-2}{3}\right)+C}$$
