Laplace of integral 0 to t (1-e^-2t) /t dt
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Step-by-step explanation:
L(1- e^-2) = 1/s - 1/s+2
Now :
L(1- e^-2) /t = integration from s to infinity (1/s - 1/s+2) ds
= [log s - log s+2] limit s to infinity
=[log (s/s+2)] limit s to infinity
=[log 1/(1+2/s)] limit s to infinity (dividing numerator and denominator by s)
above expression can be written as
= [log1 - log (1+2/s)] limit s to infinity
= -[log (1+2/s)] limit s to infinity ( as log 1= 0)
on putting limits we get
log[(s+2)/s]
Now:
L (integral 0 to t (1-e^-2t) /t dt) = 1/s × log[(s+2)/s] (ANS)
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