What is laplacetransform 1+2t ?

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

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laplacetransform 1+2t

laplace transform

1 + 2 t

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

Solution steps

L {1 + 2 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} + 2 L {t}

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

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

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

Refine

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

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

\int \frac{u}{u ^{2} + 1} d u

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

\int ln (x^{2}) d x

\int 2 x (x^{2} + 2)^{- 3} d x

i = 1 \sum \infty 8 (\frac{5}{6})^{i - 1}

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