partial fractions decomposition issue How do you decompose the following fraction?
$$\frac{2}{2x^2 (x+1)} = \frac{a}{2x^2} + \frac{b}{x+1}$$
i just want to know if I am on the right track or I am missing something(based on the above beginning).

Notice that you can cancel the common factor $2$ first.
Since $x=0$ is a double root, you need to take:
$$\frac{1}{x^2 (x+1)} = \frac{a}{x^2} + \frac{b}{x} + \frac{c}{x+1}$$
Can you take it from there?

Sometimes, you can avoid having to solve a system (in your case of three linear equations in three unknowns) to find the coefficients/numerators with a few handy manipulations.
$$\begin{align}
\frac{1}{x^2 (x+1)} 
& = \frac{1\color{blue}{+x-x}}{x^2 (x+1)} \\[5pt]
&  = \frac{1+x}{x^2 (x+1)} - \frac{x}{x^2 (x+1)} \\[5pt]
&  = \frac{1}{x^2} - \frac{1}{x (x+1)}\\[5pt]
&  = \frac{1}{x^2} - \frac{1\color{blue}{+x-x}}{x (x+1)}\\[5pt]
&  = \frac{1}{x^2} - \frac{1+x}{x (x+1)}+\frac{x}{x (x+1)}\\[5pt]
&  = \frac{1}{x^2} - \frac{1}{x}+\frac{1}{x+1}
\end{align}$$
