<s>[INST] 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?  
 [/INST] 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
$$\frac{1}{2}\left(\frac{1}{x^2+1} -\log(x^2+1)\right)+ \log(x)+c$$
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