What is inverse oflaplace (6(s+1))/(s^2(s+1)) ?

The answer to inverse oflaplace (6(s+1))/(s^2(s+1)) is 6(Undefined-\H(t)+t+e^{-t})

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inverse oflaplace (6(s+1))/(s^2(s+1))

inverse laplace

\frac{6 ( s + 1 )}{s ^{2} ( s + 1 )}

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

Solution steps

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

Use the constant multiplication property of Inverse Laplace Transform:

For function

f (t)

and constant

a : L^{- 1} {a \cdot f (t)} = a \cdot L^{- 1} {f (t)}

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

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

Undefined

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

= 6 (Undefined - H (t) + t + e^{- t})

\frac{\partial}{\partial x} (arcsin (2 x))

\int (x y) d x

y^{^{''}} + 81 y = 0, y (0) = 3, y^{^{'}} (0) = 5

x^{\frac{1}{2}}, x^{2}

y = \frac{9}{x} + 2 sin (x)

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