Integrate: $\int \frac{x}{\left(x^2-4x-13\right)^2}dx$. 
Integrate:
$$\int \frac{x}{\left(x^2-4x-13\right)^2}dx$$

Here's my attempt:
I first completed the squares for the denominator:
$$\left(x^2-4x-13\right)^2=(x-2)^2-17 \implies \int \frac{x}{\left(\left(x-2\right)^2-17\right)^2}dx$$
I then used $u$-subsituition:
$$u=x-2 \implies \int \frac{u+2}{\left(u^2-17\right)^2}du = \int \frac{u}{\left(u^2-17\right)^2}du+\int \frac{2}{\left(u^2-17\right)^2}du$$
The first part of the new integral is quite simple:
$$\int \frac{u}{\left(u^2-17\right)^2}du=\frac{-1}{2(u^2-17)}$$
Then I did the second part:
$$\int \frac{2}{\left(u^2-17\right)^2}du = -\frac{1}{2\left(u^2-17\right)}+2\left(\frac{1}{68\sqrt{17}}\ln \left|u+\sqrt{17}\right|-\frac{1}{68\left(u+\sqrt{17}\right)}-\frac{1}{68\sqrt{17}}\ln \left|u-\sqrt{17}\right|-\frac{1}{68\left(u-\sqrt{17}\right)}\right) = -\frac{1}{2\left(\left(x-2\right)^2-17\right)}+2\left(\frac{1}{68\sqrt{17}}\ln \left|x-2+\sqrt{17}\right|-\frac{1}{68\left(x-2+\sqrt{17}\right)}-\frac{1}{68\sqrt{17}}\ln \left|x-2-\sqrt{17}\right|-\frac{1}{68\left(x-2-\sqrt{17}\right)}\right) = -\frac{1}{2\left(x^2-4x-13\right)}+2\left(\frac{1}{68\sqrt{17}}\ln \left|x-2+\sqrt{17}\right|-\frac{1}{68\left(x-2+\sqrt{17}\right)}-\frac{1}{68\sqrt{17}}\ln \left|x-2-\sqrt{17}\right|-\frac{1}{68\left(x-2-\sqrt{17}\right)}\right) + C, C \in \mathbb{R}$$
Is this working out correct? I'm not really sure how WolframAlpha works, so I didn't check it on there.
 A: Here is an alternative method to integrate as follows
$$\int \frac{x}{\left(x^2-4x-13\right)^2}dx$$
$$=\int\frac12 \frac{(2x-4)+4}{\left(x^2-4x-13\right)^2}dx$$
$$=\frac12\int \frac{2x-4}{\left(x^2-4x-13\right)^2}dx+\frac12\int\frac{4}{\left(x^2-4x-13\right)^2}dx$$
$$=\frac12\int \frac{d(x^2-4x-13)}{\left(x^2-4x-13\right)^2}+2\int\frac{d(x-2)}{\left((x-2\right)^2-17)^2}$$
using reduction formula: $\color{blue}{\int \frac{dt}{(t^2+a)^n}=\frac{t}{2(n-1)a(t^2+a)^{n-1}}+\frac{2n-3}{2(n-1)a}\int\frac{dt}{(t^2+a)^{n-1}}} $,
$$=\frac12 \frac{-1}{\left(x^2-4x-13\right)}+2\left(\frac{(x-2)}{2(-17)((x-2)^2-17)}+\frac{1}{2(-17)}\int \frac{d(x-2)}{(x-2)^2-17}\right)$$
using standard formula: $\color{blue}{\int \frac{dt}{t^2-a^2}=\frac{1}{2a}\ln\left|\frac{t-a}{t+a}\right|}$,
$$=-\frac{1}{2\left(x^2-4x-13\right)}-\frac{(x-2)}{17(x^2-4x-13)}-\frac{1}{34\sqrt{17}}\ln\left|\frac{x-2-\sqrt{17}}{x-2+\sqrt{17}}\right|+C $$
$$=-\frac{2x+13}{34(x^2-4x-13)}-\frac{1}{34\sqrt{17}}\ln\left|\frac{x-2-\sqrt{17}}{x-2+\sqrt{17}}\right|+C $$
A: Hint:
Observe that
$$\left(-\frac u{u^2-a^2}\right)'=\frac{u^2+a^2}{(u^2-a^2)^2}=\frac1{u^2-a^2}+\frac{2a^2}{(u^2-a^2)^2}.$$
Hence
$$\int\dfrac{du}{(u^2-a^2)^2}=-\frac u{2a^2(u^2-a^2)}-\frac1{2a^2}\int\dfrac{du}{u^2-a^2}.$$
The last integral by $\text{artanh}$.
A: Making the problem more general
$$I=\int \frac x {(x^2+ax+b)^2}\,dx \qquad \text{with} \qquad a^2-4b \neq 0$$ Let $r$ and $s$ be the roots of the quadratic (whatever they could be - real or complex) to make
$$ \frac x {(x^2+ax+b)^2}= \frac x {(x-r)^2 \, (x-s)^2}$$ Using partial fraction decomposition
$$ \frac x {(x-r)^2 \, (x-s)^2}=\frac{r+s}{(r-s)^3}\left(\frac 1{x-s}-\frac 1{x-r} \right)+\frac 1{(r-s)^2 }\left(\frac{r}{(x-r)^2}+\frac{s}{(x-s)^2}\right)$$ The first part is simple. For the second piece, you have two integrals
$$J_k=\int \frac k {(x-k)^2}\,dx=\int \frac {dy}{(y-1)^2}=-\frac 1{y-1}=-\frac{k}{x-k}$$ Just combine all the pieces and, at the end, replace $r$ and $s$ by their values.
A: Express the second integral as
\begin{align}
\int \frac{2}{\left(u^2-17\right)^2}du
&= -\frac2{17}\int \frac{du}{u^2-17} +\frac2{17}\int \frac{u^2du}{(u^2-17)^2}
\end{align}
with
\begin{align}
\int \frac{u^2du}{(u^2-17)^2}
= -\frac12\int u d\left(\frac1{u^2-17} \right)
= -\frac u{2(u^2-17)}+\frac12\int \frac {ud u}{u^2-17}
\end{align}
Then, combining with the first integral to get
\begin{align}
\int \frac{(u+2)du}{\left(u^2-17\right)^2}=
\frac {-u}{17(u^2-17)} -\frac2{17}\int \frac{du}{u^2-17}
 & +\frac1{17}\int \frac {ud u}{u^2-17} + \int \frac{udu }{\left(u^2-17\right)^2}
\end{align}
where each piece can be integrated steadily.
