Answer
$$\dfrac{1}{s^2-2s+2}$$
Work Step by Step
The Laplace Transform can be written as:
$L[F(t)]=\int_{0}^{\infty} e^{-st} f(t) dt $
We are given that $f(t)=e^{t} \sin (t)$
Now, $L[F(t)]=\int_{0}^{\infty} e^{-st} f(t) dt \\=\int_{0}^{\infty} e^{-st} [e^{t} \sin (t)] dt\\=\int_{0}^{\infty} e^{-(s-1)t} \sin (t) \ dt \\=\dfrac{1}{s^2-2s+2}$