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

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

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simplify l^{-1}*{1/((s^2+a^2))}

simplify

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

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

Solution steps

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

Apply rule:

(a) = a

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

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

Apply exponent rule:

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

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

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

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} + a ^{2} )}

Multiply the numbers:

1 \cdot 1 = 1

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

1000000 x 1.0 8^{100}

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(\frac{1}{2 ^{x}} + \frac{1}{2 ^{x}} + \frac{1}{2 ^{x}} + \frac{1}{2 ^{x}})^{2}

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