Inverse Laplace Transformation of Inverse Tan function

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Homework Statement



Take the Inverse Laplace Transform of F(s) where
F(s)=((s)(tan-1(1/s)))


Homework Equations





The Attempt at a Solution


i know that f(t)=L-1(F(s))=(-1/t)L-1(F'(s))
and d/ds(1/tan-1(x))=1/x^2 +1
but the example I'm given with an inverse laplace of tan-1 is way prettier than this problem. hint?
 
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Well, the first thing to note is that
\tan^{-1}\left(\frac{1}{s}\right)=\cot^{-1}{(s)}.​

Do you know the convolution theorem?

Also, if you're not going to use latex, please use the "sup" and "sub" buttons (they're the buttons that look like \text{X}^2 and \text{X}_2). And make sure you use parentheses for the denominators of fractions. That said, I highly recommend you learn at least some basic latex. It's not very difficult.
 
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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