Row Echelon, Partial Fractions, and Numerator Coefficients

Question

I am trying to get the numerator values for the partial fraction decomposition of: $$\dfrac{x^2+1}{x(x-1)(x+1)(x-4)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}+\frac{D}{x-4}$$ I really started hitting speed bumps on this one, and was forced to take it slow so I decided to use a Augmented Matrix to solve the problem except that the matrix did not give the right solutions that Symbolab gives.

$$\begin{bmatrix}1&1&1&1&|&0\\-4&-3&-5&0&|&1\\ -1&-4&4&-1&|&0\\4&0&0&0&|&1\end{bmatrix}$$

This matrix does not give me the right values Symbolab gives, I would like to really learn to solve this problem using an augmented matrix it would help my skills, I also using row echelon form to solve it. If anybody has any hints, and suggestions it would greatly appreciated!

Edit

My matrix was constructed by the following technique I get a system of equations This is $x^3$ $$A+B+C+D=0$$ This is $x^2$ $$-4A-3B-5C+0D=1$$ This is $x$ $$-A-4B+4C-D=0$$ This is constants $$4A+0B+0C+0D=1$$

Answer 1

Lets use Wolfram Alpha to get a final result of the original expression using partial fractions.

We see that we would get

$$A = \dfrac{1}{4}, B = -\dfrac{1}{3}, C = -\dfrac{1}{5} , D = \dfrac{17}{60}$$

We want to solve the partial fraction expansion of

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

If we combine the RHS, we get the expression

$\dfrac{x^2+1}{x(x-1)(x+1)(x-4)} = \\ \dfrac{a x^3-4 a x^2-a x+4 a+b x^3-3 b x^2-4 b x+c x^3-5 c x^2+4 c x+d x^3-d x}{x(x-1)(x+1)(x-4)} $

Using this, we can setup four equations and four unknowns and get the augmented matrix (your matrix setup looks perfect),

$$\left[\begin{array}{rrrr|r}1&1&1&1&0\\-4&-3&-5&0&1\\ -1&-4&4&-1&0\\4&0&0&0&1\end{array}\right]$$

We perform the RREF and arrive at

$$\left[\begin{array}{rrrr|r} 1 & 0 & 0 & 0 & \dfrac{1}{4} \\ 0 & 1 & 0 & 0 & -\dfrac{1}{3} \\ 0 & 0 & 1 & 0 & -\dfrac{1}{5} \\ 0 & 0 & 0 & 1 & \dfrac{17}{60} \\ \end{array}\right]$$

If you want to see the steps of those reductions, click this link.

That matches the Wolfram result.

Where is your result failing?

The limit approach is much cleaner.

Answer 2

Denoting $$ R=\frac{X^{2}+1}{X\left(X-1\right)\left(X+1\right)\left(X-4\right)} $$

Then, since all of $ R $'s poles are simple, we have : \begin{aligned} A&=\lim_{x\to 0}{xR\left(x\right)}=\frac{1}{4}\\ B&=\lim_{x\to 1}{\left(x-1\right)R\left(x\right)}=-\frac{1}{3}\\ C&=\lim_{x\to -1}{\left(x+1\right)R\left(x\right)}=-\frac{1}{5}\\ D&=\lim_{x\to 4}{\left(x-4\right)R\left(x\right)}=\frac{17}{60} \end{aligned}