1. ## Partial Fraction Decomposition

I'm having trouble with this one problem. The answer I'm getting is apparently not correct but I can't find my error.
\begin{align} \frac{x^4+x^3+x^2-x+1}{x(x^2+1)^2}\\ =\frac{A}{x}+\frac{Bx+C}{x^2+1}+\frac{Dx+E}{(x^2+1)^2}\\ =A(x^2+1)^2+(Bx+C)(x^2+1)+(Dx+E)(x)\\ =Ax^4+2Ax^2+A+Bx^3+Bx+Cx^2+C+Dx^2+Ex\\ =(A)x^4+(B)x^3+(2A+C+D)x^2+(B+E)x+(A+C)\\ \begin{cases}A=1\\ B=1\\ 2A+C+D=1\\ B+E=-1\\ A+C=1\end{cases} \begin{cases} A=1 \\ B=1 \\ C=0\\ D=-1\\ E=-2\\ \end{cases}\\ \frac{1}{x}+\frac{x}{x^2+1}-\frac{x-2}{(x^2+1)} \end{align}

2. ## Re: Partial Fraction Decomposition

Originally Posted by Remriel
I'm having trouble with this one problem. The answer I'm getting is apparently not correct but I can't find my error.
\begin{align} \frac{x^4+x^3+x^2-x+1}{x(x^2+1)^2}\\ =\frac{A}{x}+\frac{Bx+C}{x^2+1}+\frac{Dx+E}{(x^2+1)^2}\\ =A(x^2+1)^2+(Bx+C)({\color{red}x})(x^2+1)+(Dx+E)(x)\\ =Ax^4+2Ax^2+A+Bx^3+Bx+Cx^2+C+Dx^2+Ex\\ =(A)x^4+(B)x^3+(2A+C+D)x^2+(B+E)x+(A+C)\\ \begin{cases}A=1\\ B=1\\ 2A+C+D=1\\ B+E=-1\\ A+C=1\end{cases} \begin{cases} A=1 \\ B=1 \\ C=0\\ D=-1\\ E=-2\\ \end{cases}\\ \frac{1}{x}+\frac{x}{x^2+1}-\frac{x-2}{(x^2+1)} \end{align}
This is the only thing I spotted on my first look through.Subsequent working will need to be amended.

3. ## Re: Partial Fraction Decomposition

Doh. That'll do it. Thanks.

4. ## Re: Partial Fraction Decomposition

This is a refresher for me why don't you have to multiply the denominator by the other two denominators?

5. ## Re: Partial Fraction Decomposition

Because $\displaystyle x^2+1$ is already a factor of $\displaystyle (x^2+1)^2$, so the LCD is just $\displaystyle x(x^2+1)^2$ as opposed to $\displaystyle x(x^2+1)(x^2+1)^2$