My thoughts:
$$\frac{(Ax + B)}{(x^2+1)} + \frac{(Cx + D)}{(x^2+4)} = \frac{x}{(x^2+1)(x^2+4)}$$
I combined the left terms
Set the numerator of the combined left term to "x" which is the numerator of the right term
I got
$$4Ax + Cx = x$$
I am not sure what to do next
On
I’m not entirely sure what you did to get there, but here are the steps to decomposing your partial fraction. You can compare and see where went wrong.
Starting off with$$\frac x{(1+x^2)(4+x^2)}=\frac {Ax+B}{1+x^2}+\frac {Cx+D}{4+x^2}$$We get rid of the fractions to see$$x=(Ax+B)(4+x^2)+(Cx+D)(1+x^2)$$Now, we set $x^2=-4$ to get rid of one of the expressions. Thus$$x=-3(Cx+D)\implies x=-3Cx-3D$$So $C=-\tfrac 13$ and $D=0$. Similarly, with the other expression, set $x^2=-1$ and we find that $A=\tfrac 13$ and $B=0$. Hence$$\frac x{(1+x^2)(4+x^2)}=\frac {x}{3(1+x^2)}-\frac x{3(4+x^2)}$$
You have only found the equation for $x$. The full set of equations is derived as follows:
$$(Ax+B)(x^2+4)+(Cx+D)(x^2+1)\equiv x$$
So compare coefficients of each of the powers of $x$:
$$(A+C)x^3\equiv 0\\(B+D)x^2\equiv 0\\(4A+C)x\equiv x\\(4B+D)1\equiv 0$$ Solving this gives $A=-C,B=-D$. So $$4A-A=1\implies A=\frac13\implies C=-\frac13$$ and $$4B-B=0\implies B=0\implies D=0$$ And this gives the left hand side as $$\frac{\frac13 x}{x^2+1}-\frac{\frac 13 x}{x^2+4}$$ as required.