I'm trying to solve this problem on my own and it involves simplifying the expression in the title.

In the solutions it says it's this:

$$\frac{1}{x^4+4x^2} = \frac{1}{4}\Biggl[\frac{1}{x^2}-\frac{1}{x^2+4}\Biggr]$$

But I can't for the life of me figure out where the 1/4 came from. Or exactly how they got to this answer...

  • 4
    $\begingroup$ Have you ever learnt partial fractions? This is precisely it. $\endgroup$ – Tyler6 May 19 at 5:51

We can write the fraction as, $$\frac{1}{x^4 +4x^2} = \frac{1}{4}\cdot \frac{4}{x^2(x^2+4)}$$ The fraction $\frac{4}{x^2(x^2+4)}$ can be written as.. $$\frac{1}{4}\cdot \frac{4}{x^2(x^2+4)}=\frac{1}{4}\Biggl[\frac{(x^2+4)-x^2}{x^2(x^2+4)}\Biggr]$$ $$=\frac{1}{4}\Biggl[\frac{1}{x^2}-\frac{1}{x^2+4}\Biggr]$$ Understand?


Hint: Writing $$\frac{1}{x^2(x^2+4)}=\frac{A}{x}+\frac{B}{x^2}+\frac{Cx+D}{x^2+4}$$ and compute the real numbers $$A,B,C,D$$


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