Row Echelon, Partial Fractions, and Numerator Coefficients 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
$$4A+0B+0C+0D=1$$
 A: 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}
A: 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.
