Find the partial fraction for: $\frac{1}{(x^2+1)^2(x-1)}$
By Sophia Bowman
I want to find the partial fraction of: $$\frac{1}{(x^2+1)^2(x-1)}$$Although I want this expressed through the format of finding the complex number that goes into $x$ so that I can get this equation in the form:$$\frac{1}{(x^2+1)^2(x-1)}-\frac{Ax+b}{f(x)}$$Where $f(x)$ represents the value of the denominator left from the original equation after finding the complex root.
I want to repeat this until I find the partial fractions mainly so that I have enough practice in understanding partial fractions through this method.
My approach:
I know that $(x^2+1)$ has a complex number given by: $x^2+1 = 0 \implies x = \sqrt{-1}$ Therefore $x = \pm i$
Plugging this into the equation above, I should get:$$\frac{A(\pm i )+b}{(\pm i -1)}$$Where $Ax+b = 1$
Although I'm unsure on how to proceed from here and would greatly appreciate assistance from mathematics community.
$\endgroup$1 Answer
$\begingroup$The decomposition into partial fractions has the form$$\frac{1}{(x^2+1)^2(x-1)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+1}+ \frac{Dx+E}{(x^2+1)^2}$$To obtain $A$, multiply both sides by $x-1$ and set $x=1$.
For $D$ and $E$, multiply by $(x^2+1)^2$ and set $x=i$.
Last, for $B$, multiply by $x$ and let $x\to\infty$. For $C$, you can set $x$ to any particular value which is not a pole of the fraction, e.g. $x=0$.
$\endgroup$ 3
