What is inverse oflaplace 1/(s^2)*3/((s+1)) ?

The answer to inverse oflaplace 1/(s^2)*3/((s+1)) is -3\H(t)+3t+3e^{-t}

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inverse oflaplace 1/(s^2)*3/((s+1))

inverse laplace

\frac{1}{s ^{2}} \cdot \frac{3}{( s + 1 )}

- 3 H (t) + 3 t + 3 e^{- t}

Solution steps

L^{- 1} {\frac{1}{s ^{2}} \cdot \frac{3}{( s + 1 )}}

Expand

\frac{1}{s ^{2}} \frac{3}{s + 1} : \frac{3}{s ^{3} + s ^{2}}

= L^{- 1} {\frac{3}{s ^{3} + s ^{2}}}

Take the partial fraction of

\frac{3}{s ^{3} + s ^{2}} : - \frac{3}{s} + \frac{3}{s ^{2}} + \frac{3}{s + 1}

= L^{- 1} {- \frac{3}{s} + \frac{3}{s ^{2}} + \frac{3}{s + 1}}

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

= - L^{- 1} {\frac{3}{s}} + L^{- 1} {\frac{3}{s ^{2}}} + L^{- 1} {\frac{3}{s + 1}}

L^{- 1} {\frac{3}{s}} : 3 H (t)

L^{- 1} {\frac{3}{s ^{2}}} : 3 t

L^{- 1} {\frac{3}{s + 1}} : 3 e^{- t}

= - 3 H (t) + 3 t + 3 e^{- t}

\int \frac{2 x + 1}{( x ^{2} + 4 ) ^{2} ( x + 1 ) ^{3}} d x

\frac{d}{d x} (x^{7} cos (x))

\frac{u}{u ^{2} + 6}

\int \frac{( x ^{2} - 16 ) ^{\frac{3}{2}}}{x ^{3}} d x

\int 6 x^{3} d x

What is inverse oflaplace 1/(s^2)*3/((s+1)) ?
The answer to inverse oflaplace 1/(s^2)*3/((s+1)) is -3\H(t)+3t+3e^{-t}