What is inverse oflaplace 6/((s+1)(2s^2+5s+3)) ?

The answer to inverse oflaplace 6/((s+1)(2s^2+5s+3)) is -12e^{-t}+e^{-t}*6t+12e^{-(3t)/2}

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inverse oflaplace 6/((s+1)(2s^2+5s+3))

inverse laplace

\frac{6}{( s + 1 ) ( 2 s ^{2} + 5 s + 3 )}

- 12 e^{- t} + e^{- t} \cdot 6 t + 12 e^{- \frac{3 t}{2}}

Solution steps

L^{- 1} {\frac{6}{( s + 1 ) ( 2 s ^{2} + 5 s + 3 )}}

Take the partial fraction of

\frac{6}{( s + 1 ) ( 2 s ^{2} + 5 s + 3 )} : - \frac{12}{s + 1} + \frac{6}{( s + 1 ) ^{2}} + \frac{24}{2 s + 3}

= L^{- 1} {- \frac{12}{s + 1} + \frac{6}{( s + 1 ) ^{2}} + \frac{24}{2 s + 3}}

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

= - L^{- 1} {\frac{12}{s + 1}} + L^{- 1} {\frac{6}{( s + 1 ) ^{2}}} + L^{- 1} {\frac{24}{2 s + 3}}

L^{- 1} {\frac{12}{s + 1}} : 12 e^{- t}

L^{- 1} {\frac{6}{( s + 1 ) ^{2}}} : e^{- t} \cdot 6 t

L^{- 1} {\frac{24}{2 s + 3}} : 12 e^{- \frac{3 t}{2}}

= - 12 e^{- t} + e^{- t} \cdot 6 t + 12 e^{- \frac{3 t}{2}}

\int_{0}^{1} \frac{x ^{2}}{( x + 3 ) ^{2}} d x

\int e^{10 x} \cdot 5 x d x

x = 4 y, x = y^{2} - 5

\int_{- 4}^{4} (2 x^{7} + 1) d x

(\frac{1}{y ^{2}} - \frac{9}{y ^{4}}) (y + 3 y^{3})

What is inverse oflaplace 6/((s+1)(2s^2+5s+3)) ?
The answer to inverse oflaplace 6/((s+1)(2s^2+5s+3)) is -12e^{-t}+e^{-t}*6t+12e^{-(3t)/2}