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If [itex]f(t)=K + 2cost[/itex] and F(s) = L{f(t)}, find all the real values of [itex]K[/itex] such that [itex]\int_{1}^{2}F(s)ds = 2ln5[/itex]

The attempt at a solution

So L{f(t)} = L{K} + L{2cost} = (K/s) + [2/(s^{2}+ 1)]

So [tex]\int_{1}^{2}\frac{K}{s}ds + \int_{1}^{2}\frac{2s}{s^{s}+1}ds = 2ln5 [/tex]

After integration(I used integration by substitution for the second integral) and simplification, I get K(ln2) + ln(2) = 2ln(5)

Finally, I get K = [ln 25 - ln2]/ln2

Is this correct?

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# Laplace Transform

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