Partial fraction of $\frac{s}{(s^2+2s+2)(s^2-2s+2)}$

Question

I am trying to find the partial fraction of:

$$\frac{s}{(s^2+2s+2)(s^2-2s+2)}$$

I started off with: $$\frac{A(s^2-2s+2)}{s^2+2s+2} + \frac{B(s^2+2s+2)}{s^2-2s+2}$$ After that I get the following equations: $A+B = 0$; $-2A+2B =1$

Giving: $A=-B$, and secondly $4B=1$, Hence: $A=-\frac{1}{4}$ and $B=\frac{1}{4}$

But is this correct?

Answer 1

Let $$\frac{s}{(s^2+2s+2)(s^2-2s+2)}=\frac{A(s^2+2s+2)+B(s^2-2s+2)}{(s^2+2s+2)(s^2-2s+2)}$$ The solutions to $$s^2+2s+2=0\space\text{are}\space s=-1\pm i$$ Using $s=-1+i$, we find $-1+i=B(4-4i)$, yielding $B=-\frac 14$.

Moreover, the solutions to$$s^2-2s+2=0\space\text{are}\space s=1\pm i$$ Using $s=1+i$, we get $1+i=A(4+4i)$, yielding $A=\frac 14$. Hence we can say that: $$\frac{s}{(s^2+2s+2)(s^2-2s+2)}=\frac{1}{4(s^2-2s+2)}-\frac{1}{4(s^2+2s+2)}$$

Both graphed here

Answer 2

Look for the decomposition in the form of: $$\dfrac{As+B}{(s-1)^2+1}+\dfrac{Cs+D}{(s+1)^2+1}.$$

Generally, when you have an irreducible of degree $n$ in the denominator, the numerator you need to guess need to have degree $n-1.$

Answer 3

You want numerators of degrees lower than your denominators. Let's note that $$\frac{s}{(s^2+2s+2)(s^2-2s+2)}=\frac{1}{4}(\frac{1}{s^2-2s+2}-\frac{1}{s^2+2s+2}).$$ Complex numbers are required to obtain partial fractions with linear denominators, viz. $$\frac{1}{s^2\pm 2s+2}=\frac{1}{(s\pm 1)^2+1}=\frac{i}{2}(\frac{1}{s\pm 1 +i}-\frac{1}{s\pm 1 -i}).$$

Answer 4

Since the factors in the denominator can't be factored over in the real numbers further, the best you can do is$$\frac {s}{(s^2+2s+2)(s^2-2s+2)}=\frac {As+B}{s^2+2s+2}+\frac {Cs+D}{s^2-2s+2}$$

Multiply both sides by the denominator to clear the fractions to get the equation$$s=(As+B)(s^2-2s+2)+(Cs+D)(s^2+2s+2)$$

Now let replace any occurrence of $s^2$ with $2s-2$, simplify, and repeat the process with replacing $s^2$ with $-2s-2$.

Answer 5

Denote $F(x)$ this rational function. A priori, the decomposition into partial fractions over $\mathbf R$ has the form: $$F(s)= \frac{As+B}{s^2+2s+2}+ \frac{Cs+D}{s^2-2s+2}.\tag{1}$$ We can determine the coefficients a little faster than with the basic method, if we observe that $F(s)$ is an odd function, so that $$ F(-s)= \frac{-As+B}{s^2-2s+2} + \frac{-Cs+D}{s^2+2s+2}=-F(s)=-\frac{As+B}{s^2+2s+2}- \frac{Cs+D}{s^2-2s+2}$$ By identification, we obtain that $\;-As+B=-Cs-D$, whence $$A=C,\quad B=-D.$$ Now, multiply both sides of $(1)$ by $s$ and let $s\to\infty$, we obtain the limits $$0= A+C,\enspace\text{whence }\; A=C=0.$$ Next, set $x=1$:$$F(1)=\frac1{5\cdot 1}=\frac B5 +\frac D1=\frac{4D}5,\enspace\text{whence }\enspace D=\frac14,\;B=-\frac 14.$$