# Can someone explain the steps of this Partial fraction decomposition?

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

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

• 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? Commented Nov 3, 2017 at 17:41
• @SmitShah Are you familiar with imaginary numbers? Commented Nov 3, 2017 at 17:42
• I am, but I wasn't aware we could use them here, I guess we can Commented Nov 3, 2017 at 17:44
• You don't have to: you can expand everything and then equate powers of x. Commented Nov 3, 2017 at 17:45
• @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$$x=(Ax+B)(4+x^2)+(Cx+D)(1+x^2)$$ Commented Nov 3, 2017 at 17:45

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.