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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

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2 Answers 2

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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$$\frac x{(1+x^2)(4+x^2)}=\frac {x}{3(1+x^2)}-\frac x{3(4+x^2)}$$

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  • $\begingroup$ How can x^2 be equal to -4? This is why I was confused, I thought you couldn't do this. I guess we can imaginary numbers? $\endgroup$
    – Smit Shah
    Commented Nov 3, 2017 at 17:41
  • $\begingroup$ @SmitShah Are you familiar with imaginary numbers? $\endgroup$
    – Crescendo
    Commented Nov 3, 2017 at 17:42
  • $\begingroup$ I am, but I wasn't aware we could use them here, I guess we can $\endgroup$
    – Smit Shah
    Commented Nov 3, 2017 at 17:44
  • $\begingroup$ You don't have to: you can expand everything and then equate powers of x. $\endgroup$
    – NickD
    Commented Nov 3, 2017 at 17:45
  • $\begingroup$ @SmitShah Yes, you are allowed. If you still aren’t very sure, there’s always the basic method of expanding the equation and comparing both sides$$x=(Ax+B)(4+x^2)+(Cx+D)(1+x^2)$$ $\endgroup$
    – Crescendo
    Commented Nov 3, 2017 at 17:45
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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.

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