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'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
