So, I tried to use some sort of "integration by parts" method here, but the non-conmutativity of $A$ and $A^T$ doesn't let me reach any further. I know that $\frac{d}{ds}e^{As}=Ae^{As}$ and, assuming that $A$ is invertible, I also have that$\int e^{As}ds=A^{-1}e^{As}$.
So, I would get something like (I am not so sure about the order thou): $$\int_0^t e^{sA}e^{sA^T} ds= A^{-1}e^{sA}e^{sA^T}\Big|_0^t-\int_0^t A^{-1}e^{sA}e^{sA^T}A^T ds$$
I see, that the integrals on the left and right are the same but some constants $A^{-1}$ and $A^T$, however since $A^{-1}$ doesn't conmute with $e^{sA^T}$ or $A^{T}$ with $e^{sA}$ I don't see a way where I could factor those constants to determine the integral.
Any ideas on how to proceed, or any alternative methods will e highly appreciated.
Denote the integral by $B$. Then $$ AB+BA^T=\int_0^t\frac{d}{ds}e^{sA}e^{sA^T}=e^{tA}e^{tA^T}-I. $$ This is a Sylvester equation. If no two eigenvalues of $A$ sum to zero, the equation is uniquely solved by $\operatorname{vec}(B)=(I\otimes A+A\otimes I)^{-1}\operatorname{vec}(e^{tA}e^{tA^T}-I)$. (See Wikipedia if you are not familiar with Kronecker product and vectorisation.)