Let $b : [0,+\infty) \to \mathbb R^n$ continuous and integrable. Prove there exists an unique solution of $x' = Ax + b(t)$ that tends to $0$ in $+\infty$ with $A \in M_n{(\mathbb R)}$.
I don't really see how to deal with this problem if not to integrate the equation. Is not a condition on $A$ needed?
If a condition is needed, then it would be $A$ is antisymmetric.
Skew-symmetric is the key here.
If $A$ is skew symmetric then $e^{At}$ is orthogonal, hence uniformly bounded for all $t$.
The solution is $x(t) = e^{At} (x_0 + \int_0^t e^{-A s}b(s) ds ) $.
In particular, since $t \mapsto \|e^{At} b(t)\|$ is integrable, if we choose $x_0 = - \int_0^\infty e^{-A s}b(s) ds $, we see that $x(t) \to 0$.