How do you express 1/(x^3 +4x) in partial fractions?

Mar 5, 2016

let
$\frac{1}{{x}^{3} + 4 x} = \frac{1}{x \left({x}^{2} + 4\right)} = \frac{a}{x} + \frac{b x}{{x}^{2} + 4} = \frac{a \left({x}^{2} + 4\right) + b {x}^{2}}{x \left({x}^{2} + 4\right)}$
Now
$1 = \left(a \left({x}^{2} + 4\right) + b {x}^{2}\right)$
putting x=0 in the above identity we have $4 a = 1 \implies a = \frac{1}{4}$
Again putting x=1, we have $5 a + b = 1$
putting $a = \frac{1}{4}$ in the 2nd equation $5 a + b = 1$
we get $\frac{5}{4} + b = 1 \implies b = - \frac{1}{4}$

Hence we can write
$\frac{1}{{x}^{3} + 4 x} = \frac{1}{4} \left(\frac{1}{x} - \frac{x}{{x}^{2} + 4}\right)$