How do I solve an inverse Laplace problem with an unfactorable denominator?

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i have the following problem
(2s+2)/(s^2+2s+5)
I am suppose to find the inverse laplace by putting in a form that i can find in a Laplace table, unfortunately the denominator does not factor!... not sure how to procede.
 
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Try completing the square. s2+ 2s+ 5= s2+ 2s+ 1+ 4= (s+1)2+ 4. Do you have a entry in your table that involves s2+ a2 in the denominator?
Can you write 2s+ 2 as 2(s+1)+ b?

Does your table of inverse Laplace transforms have a suggestion if you have (s+ a) instead of s?
 
yeah it does, thanks I got it.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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