What is inverse oflaplace (4s-1)/(s^2(s+1)^3) ?

The answer to inverse oflaplace (4s-1)/(s^2(s+1)^3) is 7\H(t)-t-7e^{-t}-e^{-t}*6t-(5e^{-t}t^2)/2

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inverse oflaplace (4s-1)/(s^2(s+1)^3)

inverse laplace

\frac{4 s - 1}{s ^{2} ( s + 1 ) ^{3}}

7 H (t) - t - 7 e^{- t} - e^{- t} \cdot 6 t - \frac{5 e ^{- t} t ^{2}}{2}

Solution steps

L^{- 1} {\frac{4 s - 1}{s ^{2} ( s + 1 ) ^{3}}}

Take the partial fraction of

\frac{4 s - 1}{s ^{2} ( s + 1 ) ^{3}} : \frac{7}{s} - \frac{1}{s ^{2}} - \frac{7}{s + 1} - \frac{6}{( s + 1 ) ^{2}} - \frac{5}{( s + 1 ) ^{3}}

= L^{- 1} {\frac{7}{s} - \frac{1}{s ^{2}} - \frac{7}{s + 1} - \frac{6}{( s + 1 ) ^{2}} - \frac{5}{( s + 1 ) ^{3}}}

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{7}{s}} - L^{- 1} {\frac{1}{s ^{2}}} - L^{- 1} {\frac{7}{s + 1}} - L^{- 1} {\frac{6}{( s + 1 ) ^{2}}} - L^{- 1} {\frac{5}{( s + 1 ) ^{3}}}

L^{- 1} {\frac{7}{s}} : 7 H (t)

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

L^{- 1} {\frac{7}{s + 1}} : 7 e^{- t}

L^{- 1} {\frac{6}{( s + 1 ) ^{2}}} : e^{- t} \cdot 6 t

L^{- 1} {\frac{5}{( s + 1 ) ^{3}}} : \frac{5 e ^{- t} t ^{2}}{2}

= 7 H (t) - t - 7 e^{- t} - e^{- t} \cdot 6 t - \frac{5 e ^{- t} t ^{2}}{2}

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

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

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

y \cdot \frac{d y}{d x} - x = 5 y^{2}

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

What is inverse oflaplace (4s-1)/(s^2(s+1)^3) ?
The answer to inverse oflaplace (4s-1)/(s^2(s+1)^3) is 7\H(t)-t-7e^{-t}-e^{-t}*6t-(5e^{-t}t^2)/2