Find intervals decrease and increase of a function $f(x)=\frac{\pi x}{2}-x\arctan x$. Find intervals decrease and increase of a function.
$f(x)=\frac{\pi x}{2}-x\arctan x$
$f'(x)=\frac{\pi}{2}-\arctan x-\frac{x}{1+x^2}$
$f''(x)=\frac{-2}{(1+x^2)^2}$
which is negative. What can I say from here?

Interval for which $f$ increases: $(-∞, ∞)$. Interval which $f$ decreases: none
$f$ is strictly increasing, concave downward and   $f(x)\to1$  as $x\to\infty$

Background information.
Since $f'(x)> 0$, $f''(x) < 0$, we conclude that $f$ is strictly increasing and concave downward.
interval of increasing for $f: (-∞, ∞)$, interval of decreasing for $f$: none
As $x\to\infty$ we have
$$f(x) = 1 - \frac1{3 x^2} + \frac1{5 x^4} - \frac1{7 x^6} + O\left(x^{-7}\right).$$
Hence $\lim\nolimits_{x\to\infty}f(x) = 1$.
