Proving that the solution to $f^{\prime}(x) = \frac{1}{x^{2} + (f(x))^{2}}$ is bounded above.

Question

I am given that $f:[0,\infty)\to \mathbb{R}$ is the unique solution to the ODE: $$f^{\prime}(x) = \frac{1}{x^{2} + (f(x))^{2}}$$ with $f(0)=1$ and I must prove that it is bounded.

I have already proven (using the fact that the derivative is always positive and the Mean Value Theorem) that it is monotonically increasing so is bounded below, specifically by $f(0)=1$. However, I am struggling to figure out how to show it is bounded above. One thought I had was to rearrange the ode to get $(f(x))^2$ the subject, however this doesn't seem like a valid method for some reason, I may just be overthinking it though.

Answer 1

$$\int _0^{x}f'(t) dt \leq ...$$

$f(x)\geq 1 \implies f'(x) \leq \frac 1{1+x^2} $ $$\int _0^{x}f'(t) dt \leq \int_0^x \frac {dt}{1+t^2} $$ $$f(x)-f(0)\leq \text{arctan}(x)\leq \pi/4$$