Faster method for partial fractions

Question

Is there a way to apply the "cover-up" method when solving for fractions of the following type? $$\frac {2x}{(x+1)(x^2+1)^2}$$ The long way would be; $$\frac {2x}{(x+1)(x^2+1)^2} = \frac {A}{x+1} + \frac{Bx+C}{x^2+1} + \frac{Dx+E}{(x^2+1)^2}.$$ It's easy to get $A$, which is $A = \frac{-1}{2}$. But for the rest, is there a fast trick?

Answer 1

Having obtained $A$, you can resort to complex numbers as Brad has suggested, as it will simplify things considerably by eliminating $A$,$B$ and $C$, and you will return to real numbers. $$\frac {2x}{(x+1)(x^2+1)^2}=\frac {A}{x+1} + \frac{Bx+C}{x^2+1} + \frac{Dx+E}{(x^2+1)^2}\\\Rightarrow 2x=A(x^2+1)^2+(Bx+C)(x+1)(x^2+1)+(Dx+E)(x+1) \color{blue}{\text{ Eq.(1)}}$$ Setting $x=i$ in Eq.($1$) will result in $$2i=(Di+E)(i+1)=(E-D)+(E+D)i \color{blue}{\text{ Eq.(2)}}$$ Equating the real component ($0$) and the imaginary component ($2$) in Eq.($2$) results in $E=D=1$.

Having found $A,E,D$ set $x=0$ in Eq.($1$), so that you obtain $$A+C+1=0\Rightarrow C=-1-A=-\frac{1}{2}$$ It remains to find $B$, which can be done by setting $x=1$ in Eq.(1) resulting in $$4A+4(B+C)+2(D+E)=2\Rightarrow B=2-A-C-\frac{(D+E)}{4}=\frac{1}{2}$$ Thus the partial fraction decomposition is $$\frac {2x}{(x+1)(x^2+1)^2}=-\frac {1}{2(x+1)} + \frac{x-1}{2(x^2+1)} + \frac{x+1}{(x^2+1)^2}$$