What is laplacetransform e^{-t}(t^2+1) ?

The answer to laplacetransform e^{-t}(t^2+1) is 2/((s+1)^3)+1/(s+1)

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laplacetransform e^{-t}(t^2+1)

laplace transform

e^{- t} (t^{2} + 1)

\frac{2}{( s + 1 ) ^{3}} + \frac{1}{s + 1}

Solution steps

L {e^{- t} (t^{2} + 1)}

Apply transform rule:

L {f (t)} = F (s)

then

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

For

e^{- t} (t^{2} + 1) : f (t) = (t^{2} + 1), a = - 1

= L {(t^{2} + 1)} (s + 1)

L {(t^{2} + 1)} : \frac{2}{s ^{3}} + \frac{1}{s}

= \frac{2}{( s + 1 ) ^{3}} + \frac{1}{( s + 1 )}

= \frac{2}{( s + 1 ) ^{3}} + \frac{1}{s + 1}

\frac{3}{2 3 x + 7}, at x = 6

\frac{d}{d x} (x^{2} arctan (x))

n = 0 \sum \infty \frac{1}{3 ^{n} + 1}

\frac{1}{2} x^{4} + 11 x^{3} + 30 x^{2}

\frac{2 d y}{d x} + 5 y = 4

What is laplacetransform e^{-t}(t^2+1) ?
The answer to laplacetransform e^{-t}(t^2+1) is 2/((s+1)^3)+1/(s+1)