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?  

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 $\displaystyle \mathcal{I} = -\int\frac{(u-1)}{u^2}du = +\int u^{-2}du+\int \frac{1}{u}du$
