What is inverse oflaplace ((s+1))/((s+1)^2) ?

The answer to inverse oflaplace ((s+1))/((s+1)^2) is e^{-t}

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inverse oflaplace ((s+1))/((s+1)^2)

inverse laplace

\frac{( s + 1 )}{( s + 1 ) ^{2}}

e^{- t}

Solution steps

L^{- 1} {\frac{( s + 1 )}{( s + 1 ) ^{2}}}

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {F (s - a)} = e^{a t} f (t)

For

\frac{s + 1}{( s + 1 ) ^{2}} : a = - 1, F (s) = \frac{s}{s ^{2}}

= e^{- t} L^{- 1} {\frac{s}{s ^{2}}}

L^{- 1} {\frac{s}{s ^{2}}} : 1

= e^{- t}

f (x) = \frac{8}{7 x}, at 1, 87

\frac{\partial}{\partial x} (x^{6} - 4 x^{2} y^{4} + 3 y^{2})

\frac{d}{d x} ((- x^{3} + 1) (- 2 x^{2} - 2 x))

\int \frac{x ^{3} + 3 x}{( 2 x - 2 ) ( x + 2 )} d x

y^{^{''}} + 6 y^{^{'}} + 5 y = 16 t e^{5 t}

What is inverse oflaplace ((s+1))/((s+1)^2) ?
The answer to inverse oflaplace ((s+1))/((s+1)^2) is e^{-t}