Decompose $\frac{x^2-2x+3}{(x-1)^2(x^2+4)}$, partial fraction braindead

Question

decompose $\frac{x^2-2x+3}{(x-1)^2(x^2+4)}$

the way my teacher wants us to solve is by substitution values for x,

I set it up like this:

(after setting the variables to the common denominator and getting rid of the denominator in the original equation)

$x^2-2x+3= \frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{Cx+D}{x^2+4}$

1)Let $x=1$, after plugging in for $x$ I got $2=5B$, or $B=\frac{2}{5}$

2)let the $x=0$, I get $3=-4A+4B+D$, and if I substitute B and simplify I get $D=4A-\frac{7}{5}$. Furthermore, no matter what value I plug in I still end up with two variables and cant seem to find a way to eliminate one. It seems like I have to set it up as the triple system of equations but I just dont know how to apply it here. Would really appreciate if you guys could give me a hint on how to go about it.

Answer 1

Hint : Start with $$\frac{x^2-2x+3}{(x-1)^2(x^2+4)}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{CX+D}{x^2+4}$$

Multiply with $(x-1)^2(x^2+4)$ and then insert the special values.

Answer 2

If you substitute $x=2i$, this gives

$\;\;\;-1-4i=((2Ci+D)(2i-1)^2=(-3-4i)(2Ci+D)=(-3D+8C)-(6C+4D)i$

Therefore $8C-3D=-1\;\;$ and $\;\;6C+4D=4,\;$ so

$\hspace{.6 in}16C-6D=-2\;\;$ and $\;\;9C+6D=6\implies 25C=4\implies C=\frac{4}{25}$.

Then $D=1-\frac{3}{2}C=1-\frac{6}{26}=\frac{19}{25},\;\;$ and $\;\;A=-\frac{4}{25}\;$ since $0=A+C$ from the coefficient of $x^3$.

Answer 3

After $B=\frac25$, you may obtain $C$ and $D$ similarly with

$$\frac{x^2-2x+3}{(x-1)^2}=\frac{A(x^2+4)}{x-1}+\frac{B(x^2+4)}{(x-1)^2}+{Cx+D}$$ by setting $x^2=-4$ to get $1=-8C+3D$, $2=3C+2D$, yielding $C=\frac4{25}$ and $D=\frac{19}{25}$. On the other hand, $A$ can be obtained by examining the limit $x\to \infty$ of above equation, which leads to $A=-C=-\frac4{25}$.