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

The answer to inverse oflaplace (s+1)/((s^2+1)^2) is (tsin(t))/2+1/2 (sin(t)-tcos(t))

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

inverse laplace

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

\frac{t sin ( t )}{2} + \frac{1}{2} (sin (t) - t cos (t))

Solution steps

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

Expand

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

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

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

= \frac{t sin ( t )}{2} + \frac{1}{2} (sin (t) - t cos (t))

\int_{0}^{\frac{π}{4}} 5 cos^{5} (2 x) d x

\frac{\partial}{\partial x} (\frac{x y}{x + y + z})

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

(y) \to (0) lim (3 x + 7 x y - 6 y)

x \to 1 lim (\frac{x}{∣ x ∣ - 1})

What is inverse oflaplace (s+1)/((s^2+1)^2) ?
The answer to inverse oflaplace (s+1)/((s^2+1)^2) is (tsin(t))/2+1/2 (sin(t)-tcos(t))