$(s + 1)^{- 2}$ is the Laplace transform of

t^{2}

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t^{3}

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e^{- 2 t}

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t e^{- t}

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Solution

The correct option is D $t e^{- t}$
$F (s) = \frac{1}{(s + 1)^{2}}$
So, $f (t) = L^{- 1} [\frac{1}{(s + 1)^{2}}]$
$= e^{- t} . L^{- 1} {\frac{1}{s^{2}}} = e^{- t} . t$

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