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Earlier discussion
2 older comments kept for context.
Emmanuel ugbor
Do trial and error...the one that gives you back the question is the answer
Franklin Precious
3s-1\(s+1)(s-3)=A/(s+1)+B/(s-3)
3s-1\{s+1)(s-3)=A(s-3)+B(s+1)
3s-1/(s+1)(s-3)=A(s-3)+B(s+1)/(s+1)(s-3)
denominator cancels each other
3s-1=A(s-3)+B(s+1)
If s=+3
3(3)-1=A(3-3)+B(3+1)
[A(0)=0]
9-1=4B
8=4B
B=8/4
B=2✓
If s=-1
3s-1=A(s-3)+B(s+1)
3(-1)-1=A(-1-3)+B(-1+1)
-4=-4A
A=-4/-4
A=1✓
Therefore,1/(s+1)+2/(s-3) is the answer