Root integrals are a type of integrals that frequently appear in various calculus problems.
Let's begin with this
$$ \displaystyle{ \begin{aligned}& \int \frac{ 1 }{ \sqrt{ x ^{ 2 } + a ^{ 2 } } } {\text{d}x} \ \left( \text{set } x = a \text{sinh} t \right) = & t \ = & \text{arcsinh} \frac{ x }{ a } \ = & \ln \left( x + \sqrt{ x ^{ 2 } + a ^{ 2 } } \right)\end{aligned} } $$
Use this result we compute
$$ \displaystyle{ \begin{aligned}& \int \sqrt{ x ^{ 2 } + a ^{ 2 } } {\text{d}x} \ \left( \text{Integration by parts} \right) = & x \sqrt{ x ^{ 2 } + a ^{ 2 } } - \int \frac{ x ^{ 2 } }{ \sqrt{ x ^{ 2 } + a ^{ 2 } } } {\text{d}x} \ = & x \sqrt{ x ^{ 2 } + a ^{ 2 } } - \int \sqrt{ x ^{ 2 } + a ^{ 2 } } {\text{d}x} + \int \frac{ a ^{ 2 } }{ \sqrt{ x ^{ 2 } + a ^{ 2 } } } {\text{d}x} \ \implies \int \sqrt{ x ^{ 2 } + a ^{ 2 } } {\text{d}x} = & \frac{ 1 }{ 2 } x \sqrt{ x ^{ 2 } + a ^{ 2 } } + \frac{ 1 }{ 2 } a ^{ 2 } \ln \left( x + \sqrt{ x ^{ 2 } + a ^{ 2 } } \right)\end{aligned} } $$
Given that $f$ is differentiable on $[a,b]$ and satisfies $|f'(x)| \leq M|f(x)|$ with $f(a) = 0$, where $M$ is a constant, we need to prove that $f \equiv 0$.
Proof. $\displaystyle{ \forall \alpha > 0 }$ , we have
$$ \frac{f'(x)}{\sqrt{f^2(x) + \alpha^2}} \leq M, \quad \forall x \in [a,b], $$
So (we used the result of root integral here!)
$$ \left{ \ln \left[ f(x) + \sqrt{f^2(x) + \alpha^2} \right] - M(x - a) \right}' \leq 0, \quad \forall x \in [a,b]. $$
So
$$ \ln \left[ f(x) + \sqrt{f^2(x) + \alpha^2} \right] - M(x - a) \leq \ln \alpha, \quad \forall x \in [a,b], $$
So
$$ f(x) + \sqrt{f^2(x) + \alpha^2} \leq \alpha e^{M(x-a)}, \quad \forall x \in [a,b]. $$
Let $\alpha \to 0^+$
$$ f(x) + |f(x)| \leq 0, \quad \forall x \in [a,b], $$
So
$$ f(x) \le 0, \quad \forall x \in [a,b]. $$
Similarly,
$$ f(x) \geq 0, \quad \forall x \in [a,b]. $$