Partial fraction decomposition of $\frac{x^3+x+2}{x(x^2+1)^2}$

Question

Help me solve this $$\dfrac{x^3+x+2}{x(x^2+1)^2}$$

It looked like a simple one, but became complicated in my hands because i tried it like this:

$$\dfrac{x^3+x+2}{x(x^2+1)^2}=\dfrac{A}{x}+\dfrac{Bx+C}{x^2+1}+\dfrac{Dx+E}{(x^2+1)^2} $$

multiply all sides by $x(x^2+1)^2$ to get

$x^3+x+2= A(x^2+1)+Bx(x^3+x)+C(x^2+1)+Dx^2+Ex$

group like terms:

$x^3+x+2=Bx^4+ Ax^2+Bx^2+Cx^2+Dx^2+Ex+A+C$ $x^3+x+2=Bx^4+ (A+B+C+D)x^2+Ex+A+C$

The rest seems like am in the wrong path..because I think $x^4$ seems to be misplaced.. any idea?

Answer 1

You're working with an incorrect equation.

After multiplying both sides by $x(x^2 + 1)^2$, you should have $$\begin{align} x^3+x+2 & = A(x^2+1)^2+Bx(x^3+x)+C(x^3+x)+Dx^2+Ex\\ &= Ax^4+ 2Ax^2 + A +Bx^4 + Bx^2 +Cx^3 + Cx +Dx^2 + E x \end{align}$$