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

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

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

inverse laplace

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

- H (t) + e^{t} + e^{t} t^{2}

Solution steps

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

Take the partial fraction of

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

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

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

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

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

= - H (t) + e^{t} + e^{t} t^{2}

\frac{\partial}{\partial x} (e^{x y + 1})

\frac{( 5 x - 2 )}{( x + 1 )}, 4

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

3 cos (x), 3 cos (2 x), 0 \leq x \leq π

(1 + y^{2} sin (2 x)) d x - 2 y cos^{2} (x) d y = 0

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