What is inverse oflaplace 1/((b^2*(\frac{1+x){b^2}))} ?

The answer to inverse oflaplace 1/((b^2*(\frac{1+x){b^2}))} is e^{-t}

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inverse oflaplace 1/((b^2*(\frac{1+x){b^2}))}

inverse laplace

\frac{1}{( b ^{2} \cdot ( \frac{1 + x}{b ^{2}} ) )}

e^{- t}

Solution steps

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

Use Inverse Laplace Transform table:

L^{- 1} {\frac{1}{s - a}} = e^{a t}

L^{- 1} {\frac{1}{x + 1}} = e^{- t}

= e^{- t}

x \to 1 - lim (1 - x)

\int \frac{1}{x ^{3} x ^{2} - 64} d x

\int 9 x e^{5 x} d x

6 x, 0, 2

x^{2} + 2 x + 1, - 3, 0

What is inverse oflaplace 1/((b^2*(\frac{1+x){b^2}))} ?
The answer to inverse oflaplace 1/((b^2*(\frac{1+x){b^2}))} is e^{-t}