Inverse Laplace transform Help

Icy950
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I couldn't figure the sol'n for this problem
Could somebody help?
Thanks a lot

Find the following Inverse Laplace transform

(L^(-1))*[1/(4s+1)]:frown:
 
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Icy950 said:
I couldn't figure the sol'n for this problem
Could somebody help?
Thanks a lot

Find the following Inverse Laplace transform

(L^(-1))*[1/(4s+1)]:frown:

That's almost direct from a table...

or are you doing these by hand?
 
I'm trying to do this by hand
Hopefully I can get some help
Thanks a lot
 
Well you will definitely need to show some work.

Also, you should check out LaTeX in the tutorial section. Then you can show your work as:

\frac{1}{4s+1}
 
It might be easier for you to see the solution if you first divide everything by 4:

\frac{1/4}{s+1/4}

Now it should be clear that:

L^{-1}(\frac{1/4}{s+1/4}) = \frac{1}{4}e^{\frac{1}{4}t}
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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