# Find the inverse Laplace transform of the function

Of $\displaystyle \frac{1}{4s^{2}-8s}$

I am approaching the partial fraction $$\displaystyle \frac{1}{4s\left(s-2\right)}=\frac{A}{4s}+\frac{B}{s-2},\text{ where }A=-\frac{1}{2},B=\frac{1}{8}$$

Memorizing the table, I have $$\displaystyle f(t)=-\frac{1}{8}+\frac{1}{8}e^{2t}$$

However, the answer seems to be $$\displaystyle \frac{1}{4}e^{t}\sinh{t}$$

How come?

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The two answers agree: just expand $\text{sinh}t$ –  awllower Mar 19 '13 at 4:21

$$\frac{1}{4}e^{t}\sinh{(t)} = \frac{1}{4}e^{t}\left(\frac{e^t - e^{-t}}{2}\right)= \frac{e^{2t}}{8} -\frac{1}{8} \\$$