What is laplacetransform 1+te^t ?

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

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laplacetransform 1+te^t

laplace transform

1 + t e^{t}

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

Solution steps

L {1 + e^{t} 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 {1} + L {t e^{t}}

L {1} : \frac{1}{s}

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

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

\frac{d}{d x} (sin^{2} (x) + y)

1 - 1 (t)

\int t^{3} (3 t^{2} - 4)^{\frac{5}{2}} d t

\int_{6}^{\infty} x e^{- 4 x} d x

(3 x^{2} + 3 x - 1)^{4}

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