What is laplacetransform t+sin(t) ?

The answer to laplacetransform t+sin(t) is 1/(s^2)+1/(s^2+1)

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laplacetransform t+sin(t)

laplace transform

t + sin (t)

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

Solution steps

L {t + sin (t)}

Use the linearity property of Laplace Transform:

For functions

f (t), g (t)

and constants

a, b : L {a \cdot f (t) + b \cdot g (t)} = a \cdot L {f (t)} + b \cdot L {g (t)}

= L {t} + L {sin (t)}

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

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

= \frac{1}{s ^{2}} + \frac{1}{s ^{2} + 1}

sin (2 t) cos (5 t)

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

\int 8 ln (x^{2} - 1) d x

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

f (x) = \frac{1}{x ln ( x )}

What is laplacetransform t+sin(t) ?
The answer to laplacetransform t+sin(t) is 1/(s^2)+1/(s^2+1)