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?
 A: Since you selected $$x - 2 = 3 \tan \theta$$ as your substitution, it follows that $$\tan \theta = \frac{x-2}{3},$$ and by considering a right triangle with legs $3$ and $x-2$ with hypotenuse $\sqrt{3^2 + (x-2)^2}$ via the Pythagorean theorem, we obtain $$\sin \theta = \frac{x-2}{\sqrt{3^2 + (x-2)^2}}, \\ \cos \theta = \frac{3}{\sqrt{3^2 + (x-2)^2}}.$$  Therefore, $$\frac{1}{2} \sin 2\theta = \sin \theta \cos \theta = \frac{3(x-2)}{3^2 + (x-2)^2} = \frac{3(x-2)}{x^2 - 4x + 13}.$$  We also easily have $$\theta = \tan^{-1} \frac{x-2}{3}.$$  Therefore $$\frac{1}{54}\left( \theta + \frac{1}{2} \sin 2\theta \right) = \frac{1}{54} \left( \tan^{-1} \frac{x-2}{3} + \frac{3(x-2)}{x^2 - 4x + 13} \right).$$
A: 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
$$\int \dfrac{dx}{(x^2-4x+13)^2}$$$$=\int \dfrac{d(x-2)}{((x-2)^2+3^2)^2}$$
$$=\frac{x-2}{2\cdot 3^2((x-2)^2+3^2)}+\frac{1}{2\cdot 3^2}\int \frac{d(x-2)}{(x-2)^2+3^2}$$
$$=\frac{x-2}{18(x^2-4x+13)}+\frac{1}{18}\left(\frac13\tan^{-1}\left(\frac{x-2}{3}\right) \right)+C$$
$$=\bbox[15px,#ffd,border:1px solid green]{\frac{x-2}{18(x^2-4x+13)}+\frac{1}{54}\tan^{-1}\left(\frac{x-2}{3}\right)+C}$$
A: 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:
$$\int \dfrac{dx}{(x^2-4x+13)^2} = \frac 1{27}I(u)= \frac 1{54}\left(\arctan \frac{x-2}{3} + \frac{\frac{x-2}{3}}{\left(\frac{x-2}{3}\right)^2+1}\right) (+C)$$
$$= \frac 1{54}\left(\arctan \frac{x-2}{3} + \frac{3(x-2)}{\left(x-2\right)^2+9}\right)(+C)$$
A: substitute $\theta=\tan^{-1}\left(\frac{x-2}{3}\right)$ & $$\sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}=\frac{2\left(\frac{x-2}{3}\right)}{1+\left(\frac{x-2}{3}\right)^2}=\frac{6(x-2)}{x^2-4x+13}$$
After substituting $\theta$ and $\sin2\theta$, you will get final answer
$$I=\frac{1}{54}\tan^{-1}\left(\frac{x-2}{3}\right)+\frac{x-2}{18(x^2-4x+13)}+C$$
A: Here is an alternative method. You can first use substitution to simplify the integral:
$$\int \dfrac{dx}{(x^2-4x+13)^2}$$
$$ =\int \dfrac{dx}{((x-2)^2+9)^2} = \int \dfrac{du}{(u^2+9)^2} \tag{$u=x-2, du = dx$}$$
$$=\frac{1}{81}\int \dfrac{du}{(\frac{u^2}{9}+1)^2} = \frac{1}{27} \int \dfrac{dv}{(v^2+1)^2} \tag{$u=3v, du = 3 \ dv$}$$
and then use integration by parts as shown in Rene's answer in this thread. This leads to:
$$2\int \frac{1}{(1+v^2)^2}\,dv=\frac{v}{1+v^2}+\int \frac{1}{1+v^2}\,dv$$
$$\int \frac{1}{(1+v^2)^2}\,dv=\frac{1}{2} \left(\frac{v}{1+v^2}+\arctan(v) \right)$$
and then a couple of back-substitutions leads you to the answer of the original problem.
A: 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
$$I=(a-b)^3 \int [1-2/v+1/v^2] dv= (a-b)^{-3}[v-2 \ln v -1/v] $$
$$I=(a-b)^{-3}\left(\frac{a-b}{x-a}-2\ln \frac{a-b}{x-a}-\frac{x-a}{a-b}\right)$$
A: By setting
$$
\frac{1}{\left(x^2-4 x+13\right)^2}=\frac{A (2 x-4)+B}{x^2-4 x+13}+\frac{d}{dx}\left(\frac{C x+D}{x^2-4 x+13}\right)
$$
you get
\begin{align}
A &= 0,\\
B &= \frac{1}{18},\\
C &= \frac{1}{18},\\
D &= -\frac{1}{9}
\end{align}
so that
\begin{align}
\int\frac{1}{\left(x^2-4 x+13\right)^2}dx 
  &= \frac{1}{18}\int\frac{1}{x^2-4 x+13}dx+\frac{x-2}{18(x^2-4 x+13)}=\\
  &= \frac{1}{54} \arctan\left(\frac{x-2}{3}\right)+\frac{x-2}{18(x^2-4x+13)}+c
\end{align}
A: 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}
A: $\begin{aligned} I&=\int \frac{d x}{\left[(x-2)^{2}+9\right]^{2}} \\ &=\int \frac{d y}{\left(y^{2}+9\right)^{2}}, \text { where } y=x-2 \\ &=-\frac{1}{2} \int \frac{1}{y} d\left(\frac{1}{y^{2}+9}\right) \quad \text{(By IBP)}\\ &--\frac{1}{2 y\left(y^{2}+9\right)}-\frac{1}{2} \int\left(\frac{1}{y^{2}} \cdot \frac{1}{y^{2}+9}\right) d y \\ &=-\frac{1}{2 y\left(y^{2}+9\right)}-\frac{1}{18} \int\left(\frac{1}{y^{2}}-\frac{1}{y^{2}+9}\right) d y \\ &=-\frac{1}{2 y\left(y^{2}+9\right)}+\frac{1}{18 y}+\frac{1}{54} \tan ^{-1}\left(\frac{y}{3}\right)+C \\ &=\frac{y}{18\left(y^{2}+9\right)}+\frac{1}{54} \tan ^{-1}\left(\frac{y}{3}\right)+C \\ &=\frac{1}{54}\left(\frac{3(x-2)}{x^{2}-4 x+13}+\tan ^{-1}\left(\frac{x-2}{3}\right) \right)+C \end{aligned}$
