Answer
See below
Work Step by Step
Since the given function is periodic on $[0,1)$ with period $T=1$, we have
$L(f)=\frac{1}{1-e^{-s}}\int ^1_0e^{-st}f(t)dt=\frac{1}{1-e^{-s}}\int ^1_0 e^t e^{-st} dt$
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
$L(f)=\frac{1}{1-e^{-s}}\int ^1_0e^t e^{-st} dt\\
=\frac{1}{1-e^{-s}}[\frac{e^{t(1-s)}}{1-s}]^1_0\\
=\frac{1-e^{1-s}}{(1-e^{-s})(1-s)}$
