What is laplacetransform 1/((8^2+1)+(8^2+4)) ?

The answer to laplacetransform 1/((8^2+1)+(8^2+4)) is 1/(133s)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

laplacetransform 1/((8^2+1)+(8^2+4))

laplace transform

\frac{1}{( 8 ^{2} + 1 ) + ( 8 ^{2} + 4 )}

\frac{1}{133 s}

Solution steps

L {\frac{1}{( 8 ^{2} + 1 ) + ( 8 ^{2} + 4 )}}

Use Laplace Transform table:

L {a} = \frac{a}{s}

L {\frac{1}{8 ^{2} + 1 + 8 ^{2} + 4}} = \frac{\frac{1}{8 ^{2} + 1 + 8 ^{2} + 4}}{s}

= \frac{\frac{1}{8 ^{2} + 1 + 8 ^{2} + 4}}{s}

= \frac{1}{133 s}

\int 2 e^{- 0.4 x} d x

\frac{d}{d x} (144 + x^{2})

\int (sin (x))^{2} (cos (x))^{2} d x

(2005, 54), (2010, 58)

3 x^{- 1}

What is laplacetransform 1/((8^2+1)+(8^2+4)) ?
The answer to laplacetransform 1/((8^2+1)+(8^2+4)) is 1/(133s)