Decomposition of this partial function I came across this 
$$\int \frac{dx}{x(x^2+1)^2}$$
in "Method of partial functions" in my Calculus I book. 
The thing is that, however, decomposing $\frac{1}{x(x^2+1)^2}$ would take long in this way: 
$$\frac{1}{x(x^2+1)^2}= \frac{A}{x}+\frac{Bx+C}{x^2+1}+\frac{Dx+E}{(x^2+1)^2}$$
So is there a quicker or a more practical way that I can use?  
 A: Let's take the decomposition you give and see how quick we can go
$$\frac{1}{x(x^2+1)^2}= \frac{A}{x}+\frac{Bx+C}{x^2+1}+\frac{Dx+E}{(x^2+1)^2}$$
Multiplying by $x$ the two sides and making $x=0$ one gets $A=1$.
Multiplying by $(x^2+1)^2$ and making $x=i$ we get $-i=Di+E$ and this means $D=-1$ and $E=0$ . Now rewrite the decomposition
$$\frac{1}{x(x^2+1)^2}= \frac{1}{x}+\frac{Bx+C}{x^2+1}-\frac{x}{(x^2+1)^2}$$
This is equivalent to
$$\begin{align}{Bx+C\over x^2+1}=&{1\over x(x^2+1)^2}-{1\over x}+{x\over (x^2+1)^2}\\=&{(x^2+1)(1-1-x^2)\over x(x^2+1)^2}\\=&-{x\over x^2+1}\end{align}$$
Looks pretty quick to me
A: One way to simplify this a bit, is to multiply numerator and denominator of your integrand by $x$:
$$\int \frac{x}{x^2(x^2+1)^2}dx$$
Now substituting $y = x^2$; $dy = 2x dx$, you get the integral 
$$\frac{1}{2}\int \frac{1}{y(y+1)^2}dy$$
Splitting this in partial fractions is more straightforward than what you had before:
$$\frac{1}{y(y+1)^2} = -\frac{1}{(y+1)^2} -\frac{1}{y+1} + \frac{1}{y}$$
Solving the integral then gives
$$\frac{1}{2}\left(\frac{1}{y+1}  + \log(y) -\log(y+1)\right)+c$$
Subsituting $y = x^2$ then gives
$$\frac{1}{2}\left(\frac{1}{x^2+1}  + \log(x^2) -\log(x^2+1)\right)+c$$
which can be slightly simplified to
$$\frac{1}{2}\left(\frac{1}{x^2+1} -\log(x^2+1)\right)+ \log(x)+c$$
A: Let $\displaystyle \mathcal{I} = \int\frac{1}{x(x^2+1)^2}dx$
substitute $\displaystyle x = \frac{1}{t}$ and $\displaystyle dx = -\frac{1}{t^2}$
So $\displaystyle \mathcal{I} = -\int\frac{t^5}{(1+t^2)^2}\cdot \frac{1}{t^2}dt = -\frac{1}{2}\int\frac{t^2\cdot 2t }{(1+t^2)^2}dt$
put $1+t^2=u\;,$ then $2tdt = du$
So $\displaystyle \mathcal{I} = -\int\frac{(u-1)}{u^2}du = +\int u^{-2}du+\int \frac{1}{u}du$
A: \begin{equation}
\int \frac{dx}{x(x^2+1)^2}
\end{equation}
\begin{equation}
\frac{1}{x(x+i)^2(x-i)^2}=\dfrac{A}{x}+\dfrac{B}{x+i}+\dfrac{C}{x-i}+\dfrac{D}{(x+i)^2}+\dfrac{E}{(x-i)^2}
\end{equation}


*

*Let $x=0$. Then $A=\dfrac{1}{(0+i)^2(0-i)^2}=1$\

*Let $x=-i$. Then $B=\dfrac{1}{-i(-i-i)^2}=-\dfrac{1}{2}$\

*Let $x=i$. Then $C=\dfrac{1}{i(i+i)^2}=-\dfrac{1}{2}$\


Therefore we have
\begin{eqnarray}
\frac{1}{x(x^2+1)^2}&=&\dfrac{1}{x}-\dfrac{1}{2}\left(\dfrac{1}{x+i}+\dfrac{1}{x-i}\right)+\dfrac{D}{(x+i)^2}+\dfrac{E}{(x-i)^2}\\
&=&\dfrac{1}{x}-\dfrac{x}{x^2+1}+\dfrac{D(x^2-2ix-1)+E(x^2+2ix-1)}{(x^2+1)^2}\\
&=&\dfrac{1}{x}-\dfrac{x}{x^2+1}+\dfrac{(D+E)x^2+2(E-D)ix-(D+E)}{(x^2+1)^2}\\
&=&\dfrac{1}{x}-\dfrac{x}{x^2+1}+\dfrac{2Eix}{(x^2+1)^2}
\end{eqnarray}
Note that since the $x^2$ term must vanish it must be the case that $D=-E$.
Thus we only have to find the value of $E$.
\begin{eqnarray}
\frac{1}{x(x^2+1)^2}&=&\dfrac{1}{x}-\dfrac{x}{x^2+1}+\dfrac{2Eix}{(x^2+1)^2}
\end{eqnarray}
This equation holds for every value of $x$ with the exception of $x=0,i,-i$. Therefore, when $x=1$ it is true that
\begin{eqnarray}
\frac{1}{4}&=&1-\dfrac{1}{2}+\dfrac{E}{2}i\\
E&=&\dfrac{i}{2}
\end{eqnarray}
Therefore
\begin{equation}
\frac{1}{x(x^2+1)^2}=\dfrac{1}{x}-\dfrac{x}{x^2+1}-\dfrac{x}{(x^2+1)^2}
\end{equation}
A: This is not a complete answer, but rather a helpful observation.
First note that the integrand is odd.
Secondly, note that for real factorisations, there is no conjugation from negative terms that can exist (e.g. $1-x$ and $1+x$)
With this in mind, we proceed as follows:
$$\frac{1}{x(x^2+1)^2} = \frac{a}{x} + \frac{bx}{x^2+1}+\frac{cx}{(x^2+1)^2}$$
This reduces the $5$ variable problem into a $3$ variable problem, which is much more manageable.
