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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

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

\frac{d x}{d t} = \frac{( e ^{t} - e ^{- t} )}{3 + x}, x (0) = 0

y = x e^{2 x}

\frac{\partial}{\partial x} (ln (x^{2} + y^{2} + 6))

\int (5 x^{5} + 3) x^{2} d x

\frac{d y}{d x} = (5 y)

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