Rational bounds for $\tan^{-1} x$ I want help with this question.

Show that for all $x>0$, $$ \frac{x}{1+x^2}<\tan^{-1}x<x.$$

Thank you.

$$ \frac{d}{dx} \tan^{-1} x  = \frac{1}{1+x^2} < \frac{d}{dx} x =1$$
and $\tan^{-1}(0)=0$ so that the last inequality is proved. 
$$  \frac{d}{dx} \frac{x}{x^2+1} = \frac{(x^2+1) - x(2x) }{(x^2+1)^2} = \frac{1-x^2}{(x^2+1)^2} < \frac{d}{dx} \tan^{-1} x  = \frac{1}{1+x^2} $$
and $\frac{x}{1+x^2}(0)=0=\tan^{-1}(0)$ so that the first inequality is proved. 
Another way : 
$$\frac{x}{x^2+1}< t\frac{1}{t^2+1}|_{t=0}^{t=x} - 
\int_0^x t\frac{-2t}{(t^2+1)^2} = \int_0^x \frac{1}{t^2+1}\ dt =\tan^{-1}x < \int_0^x \frac{1}{0+1}\ dt=x$$
