Answer:
[tex]\frac{1/3}{s^2+1}- \frac{1/3}{s^2+4}[/tex]
Step-by-step explanation:
You already have the bottom in factors, write the partial fraction in the form As+b for each factor
[tex]\frac{1}{(s^2+1)(s^2+4)} =\frac{As+B}{s^2+1}+ \frac{Cs+D}{s^2+4}[/tex]
Multiply both sides by the bottom to stop using fractions:
[tex]\frac{1}{(s^2+1)(s^2+4)}(s^2+1)(s^2+4)=(\frac{As+B}{s^2+1}+ \frac{Cs+D}{s^2+4})(s^2+1)(s^2+4)[/tex]
[tex]1=(As+B)(s^2+4)+(Cs+D)(s^2+1)}[/tex]
To find the value of the constants A, B, C, and D, separate the equation by the grade of the s, for example, all that multiplies s^3 in one equation like this:
[tex]0=As^3+Cs^3[/tex]
Because in the left part of the equation doesn't have constants multiplying the s^3 we put a zero in that side of the equation, do the same with s^2,s^1, and s^0
[tex]0=4As+Cs\\0=Bs^2+Ds^2\\1=4B+D[/tex]
In this case, you can group the equation by constant to solve. First A and D, then B and C.
[tex]As^3+Cs^3=0\\4As+Cs=0[/tex]
[tex]Bs^2+Ds^2=0\\4B+D=1[/tex]
Solving for A and C:
[tex](A+C)(s^3)=0\\(4A+C)s=0[/tex]
[tex]A+C=0\\4A+C=0\\\\[/tex]
This system only has one solution A=0 and C=0.
Solving for B and D:
[tex](B+D)(s^2)=0\\4B+D=1[/tex]
[tex]\\B+D=0\\4B+D=1\\\\D=-B\\4B-B=1\\3B=1\\B=1/3\\D=-1/3[/tex]
Then the solution is:
[tex]\frac{1/3}{s^2+1}- \frac{1/3}{s^2+4}[/tex]
