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

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

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

inverse laplace

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

e^{- t} t

Solution steps

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

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {F (s - a)} = e^{a t} f (t)

For

\frac{1}{( t + 1 ) ^{2}} : a = - 1, F (t) = \frac{1}{t ^{2}}

= e^{- t} L^{- 1} {\frac{1}{t ^{2}}}

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

= e^{- t} t

\int \frac{x + 4}{2 x + 3} d x

y^{^{''}} + y = e^{x} sin (x)

\int \frac{2 x}{x} d x

\int 1 + 2 cos (x) + cos^{2} (x) d x

y = x^{3}, y = x^{2}

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