partial fractions decomposition issue

Question

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).

Answer 1

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?

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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.

$$\begin{align} \frac{1}{x^2 (x+1)} & = \frac{1\color{blue}{+x-x}}{x^2 (x+1)} \\[5pt] & = \frac{1+x}{x^2 (x+1)} - \frac{x}{x^2 (x+1)} \\[5pt] & = \frac{1}{x^2} - \frac{1}{x (x+1)}\\[5pt] & = \frac{1}{x^2} - \frac{1\color{blue}{+x-x}}{x (x+1)}\\[5pt] & = \frac{1}{x^2} - \frac{1+x}{x (x+1)}+\frac{x}{x (x+1)}\\[5pt] & = \frac{1}{x^2} - \frac{1}{x}+\frac{1}{x+1} \end{align}$$