Proof using Gronwall inequality

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Using the Gronwall inequality, how do I show that if:

$$\partial_t \log (1 + u(t) ^2) <1$$

Such that $u(t)$ is a solution of the differential equation :

$$\dot x(t) = - ax(t)+ \log (1+x(t)^2)\quad(a>1) $$

Then any $u(t)$ tends to $0$ when $t\to+\infty $.

Remark: I already proved any $u(t)$ is defined for all real $t$.

I tried to show that when $t\to+ \infty $ , $\log (1 + u(t) ^2) \leq 0$ and therefore $u(t)$ must tend to zero.