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

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

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

inverse laplace

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

e^{- t} t

Solution steps

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

Factor

s^{2} + s^{1} + s^{1} + 1 : (s + 1)^{2}

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

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

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

= e^{- t} t

\int \frac{1}{u 5 - u ^{2}} d u

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

\int_{0}^{r} x r^{2} - x^{2} d x

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