how to compute this limits given these conditions. 
if $f(1)=1$ and $f'(x)=\frac{1}{x^2+[f(x)]^2}$ then compute $\lim\limits_{x\to+\infty}f(x)$

i tried to write it was
$$\frac{dy}{dx}=\frac{1}{x^2+y^2}\\
(x^2+y^2)\frac{dy}{dx}=1\\
(x^2+y^2)dy=dx$$
by the help
$$\begin{align}
f(x)&\le1+\int_1^x\frac{dt}{1+t^2}\\
&\le1+\arctan t\bigg|_1^x\\
&\le1+\arctan x-\arctan 1\\
&\le1+\arctan x-\frac{\pi}{4}
\end{align}$$
so
$$\begin{align}
\lim\limits_{x\to+\infty}f(x)&\le\lim\limits_{x\to+\infty}1+\arctan x-\frac{\pi}{4}\\
&\le1+\frac{\pi}{2}-\frac{\pi}{4}\\
&\le1+\frac{\pi}{4}=\frac{4+\pi}{4}
\end{align}$$

Since $f'(x)>0 $ and $f(1)=1$ then we have 
$$ \frac{1}{x^2+f(x)^2}\leq \frac{1}{x^2+1}. $$
Also we will have 
$$ \int_{1}^{x}f'(t)dt = \int_{1}^{x}\frac{dt}{t^2+f(t)^2} \leq \int_{1}^{x}\frac{dt}{t^2+1} $$
$$ \implies f(x) \leq 1+ \int_{1}^{x}\frac{dt}{t^2+1} . $$
Try to finish the problem.
