<s>[INST] Inverse Laplace Transform of $ \left(\frac{1-s^{1/2}}{s^2}\right)^2$ I found this question in my N.P Bali's Engineering Mathematics 7th Edition.
I could not find any solved questions related to this. 
How can I find the Inverse Laplace Transform of : 
$\left({1-s^{1/2} \over s^{2}}\right)^2$
I know only to find the inverse Laplace of $s$ when the power is a whole number. 
The answer to the question is given as : $\dfrac{t^3}{6} + \dfrac{t^2}{2} - \dfrac{16t^\frac{5}{2}}{15 \pi^{\frac{1}{2}}}$
I want to know how to get it. 
 [/INST] The partial fraction expansion yields;
$$\left(\dfrac{1-s^{1/2}}{s^{2}}\right)^2 = -\dfrac{2}{s^{7/2}}+\dfrac{1}{s^4}+\dfrac{1}{s^3} $$
Using this table of Laplace Transforms (item $6$ and item $3$ (twice)) yields:
$$\mathscr{L}^{-1} \left(-\dfrac{2}{s^{7/2}}+\dfrac{1}{s^4}+\dfrac{1}{s^3}\right) = -\dfrac{2 \times 2^3}{1 \times 3 \times 5 ~ \sqrt{\pi}}t^{5/2} +\dfrac{t^3}{6} + \dfrac{t^2}{2} = -\dfrac{16}{15~ \sqrt{\pi}}t^{5/2} +\dfrac{t^3}{6} + \dfrac{t^2}{2}$$
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