What is l^{-1}*[ 1/((s^2+4s+4))] ?

The solution to l^{-1}*[ 1/((s^2+4s+4))] is 1/(l(s^2+4s+4))

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simplify l^{-1}*[ 1/((s^2+4s+4))]

simplify

l^{- 1} \cdot [\frac{1}{( s ^{2} + 4 s + 4 )}]

\frac{1}{l ( s ^{2} + 4 s + 4 )}

Solution steps

l^{- 1} (\frac{1}{s ^{2} + 4 s + 4})

Apply rule:

(a) = a

(\frac{1}{s ^{2} + 4 s + 4}) = \frac{1}{s ^{2} + 4 s + 4}

= l^{- 1} \frac{1}{s ^{2} + 4 s + 4}

Apply exponent rule:

a^{- 1} = \frac{1}{a}

l^{- 1} = \frac{1}{l}

= \frac{1}{l} \cdot \frac{1}{s ^{2} + 4 s + 4}

Apply the fraction rule:

\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}

= \frac{1 \cdot 1}{l ( s ^{2} + 4 s + 4 )}

Multiply the numbers:

1 \cdot 1 = 1

= \frac{1}{l ( s ^{2} + 4 s + 4 )}

v (\frac{( 1 - cos ( x ) )}{( 1 + cos ( x ) )})

m^{3} + 27

(7 + m)^{3}

(\frac{2 x - 1}{2 \cdot y - 3 z})^{2}

7^{2 x} - 2 (7^{x^{2} + x + 12}) + 7^{2 x + 24}

What is l^{-1}*[ 1/((s^2+4s+4))] ?
The solution to l^{-1}*[ 1/((s^2+4s+4))] is 1/(l(s^2+4s+4))