Good afternoon,
It was just to see if the book was wrong where I had put in the box. Shouldn't it be $e^{-u}$ when replacing the antitransformed one?
By the way $f(t)=e^t$
$$y(t)=\mathcal{L}^{-1}\left[ F(p)\circ G(p)\right] =\int_0^t f(t-u)\cdot g(u)\cdot du,$$ $$\text{non}~\mathcal{L}^{-1}\left[ \dfrac{1}{(p+1)^2+1} \right]=e^{-t}\cdot \sin t\cdot dt\Rightarrow y(t)=\int_0 ^t e^{-(t-u)}\cdot \boxed{e^u}\cdot \sin (u)\cdot du$$