What is l^{-1}(1/((s+3)^n)) ?

The solution to l^{-1}(1/((s+3)^n)) is 1/(l(s+3)^n)

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simplify l^{-1}(1/((s+3)^n))

simplify

l^{- 1} (\frac{1}{( s + 3 ) ^{n}})

\frac{1}{l ( s + 3 ) ^{n}}

Solution steps

l^{- 1} (\frac{1}{( s + 3 ) ^{n}})

Apply rule:

(a) = a

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

= l^{- 1} \frac{1}{( s + 3 ) ^{n}}

Apply exponent rule:

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

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

= \frac{1}{l} \cdot \frac{1}{( s + 3 ) ^{n}}

Apply the fraction rule:

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

= \frac{1 \cdot 1}{l ( s + 3 ) ^{n}}

Multiply the numbers:

1 \cdot 1 = 1

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

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What is l^{-1}(1/((s+3)^n)) ?
The solution to l^{-1}(1/((s+3)^n)) is 1/(l(s+3)^n)