I'd like to prove
If $u:[0,a]\rightarrow \mathbb{R}$ is continuous and nonnegative and $u(t)\leq c+\int_{0}^{t}[Ku(s)+M]ds$, then $u(t)\leq ce^{Kt}+\frac{M}{K}(e^{Kt}-1)$ where $c>0$, $K>0$, and $M\geq0$.
from
If $u,k:[0,a]\rightarrow \mathbb{R}$ are continuous, $k$ is nonnegative and $u(t)\leq c+\int_{0}^{t}[k(s)u(s)]ds$, then $u(t)\leq ce^{\int_{0}^{t}k(s)ds}$.
$ u(t)\leq c+K\int^t_0(u+\frac{M}{K})\,ds$ implies $$ u(t)+\frac{M}{K}\leq c+\frac{M}{K} + K\int^t_0(u+\frac{M}{K})\,ds$$ Let $v(t)= u(t)+\frac{M}{K}$. Then
$$v(t)\leq h(t)=c+ \frac{M}{K} + K\int^t_0 v(s)\,ds$$
$\dot{h}(t)=K v\leq Kh$
Then $\dot{h}-Kh\leq 0$. Multiply the integrating factor $e^{Kt}$ tu get
$$ \Big(e^{-Kt}h\Big)'\leq 0 $$
HEnce
$$ e^{-Kt}h(t)-(c+\frac{M}{K}) \leq 0 $$ From where $$ u(t)+\frac{M}{K}\leq h(t)\leq (c+\frac{M}{K})e^{Kt}$$