What is inverse oflaplace 8/((s+1)^2) ?

The answer to inverse oflaplace 8/((s+1)^2) is e^{-t}*8t

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inverse oflaplace 8/((s+1)^2)

inverse laplace

\frac{8}{( s + 1 ) ^{2}}

e^{- t} \cdot 8 t

Solution steps

L^{- 1} {\frac{8}{( 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{8}{( s + 1 ) ^{2}} : a = - 1, F (s) = \frac{8}{s ^{2}}

= e^{- t} L^{- 1} {\frac{8}{s ^{2}}}

L^{- 1} {\frac{8}{s ^{2}}} : 8 t

= e^{- t} \cdot 8 t

\frac{d ^{2} x}{d t ^{2}} + \frac{d x}{d t} + x = 0.5 sin (2 t)

\int \frac{1}{x ^{2} - 25} d x

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

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

x \to 1 lim (\frac{( x ^{2} - 5 )}{x})

What is inverse oflaplace 8/((s+1)^2) ?
The answer to inverse oflaplace 8/((s+1)^2) is e^{-t}*8t