Answer
See below
Work Step by Step
Since the given function is periodic on $[0,\infty)$ with period $T=2$, we have
$L(f)=\frac{1}{1-e^{-2s}}\int ^2_0e^{-st}f(t)dt$
Using the standard integral
$\int e^{at}\sin bt dt=\frac{1}{a^2+b^2}e^{at}(a\sin bt-b\cos bt)+c$
it follows that
$L(f)=\frac{1}{1-e^{-2 s}}[\int ^1_0 e^{-st} dt-\int^2_1e^{-st} dt]\\
=\frac{1}{1-e^{-2 s}}[(\frac{e^{-st} }{-s}|^1_0)- (\frac{e^{-st}}{s}dt |^2_1)]\\
=\frac{1}{1-e^{-2 s}}[\frac{1-e^{-st}}{s}+\frac{e^{-2s}-e^{-s}}{s}]\\
=\frac{(e^{-s}-1)^2}{s(1-e^{-2 s})}$
