What is inverse oflaplace (s+1)/((s+1)^2+5) ?

The answer to inverse oflaplace (s+1)/((s+1)^2+5) is e^{-t}cos(sqrt(5)t)

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inverse oflaplace (s+1)/((s+1)^2+5)

inverse laplace

\frac{s + 1}{( s + 1 ) ^{2} + 5}

e^{- t} cos (5 t)

Solution steps

L^{- 1} {\frac{s + 1}{( s + 1 ) ^{2} + 5}}

Apply inverse transform rule:

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

then

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

For

\frac{s + 1}{( s + 1 ) ^{2} + 5} : a = - 1, F (s) = \frac{s}{s ^{2} + 5}

= e^{- t} L^{- 1} {\frac{s}{s ^{2} + 5}}

L^{- 1} {\frac{s}{s ^{2} + 5}} : cos (5 t)

= e^{- t} cos (5 t)

y = x^{5} \cdot x^{4}

\frac{d}{d x} (6 sin (x) - 6 x cos (x))

y^{^{'}} + x y^{^{''}} = 0

\int cos^{2} (n x) d x

\int_{0}^{1} (x^{2} + 7) e^{- x} d x

What is inverse oflaplace (s+1)/((s+1)^2+5) ?
The answer to inverse oflaplace (s+1)/((s+1)^2+5) is e^{-t}cos(sqrt(5)t)