Proving that the integral $\int_1^2 \frac{\arctan(x)}{x} \,dx$ is less than $1$ I am looking to prove that:
$$\int_1^2 \frac{\arctan(x)}{x} \,dx< 1$$
Using Darboux higher sum i can do that pretty easily, however, in order to use Darboux sum (which is given by plugging the value $1$ to the function) I also need to show that the function is descreasing, and things got a little messed up there..
any help?
 A: HINT:
$$\int_1^2 \frac{\arctan(x)}{x}\,dx\le \int_1^2 \arctan(x)\,dx$$
or 
$$\int_1^2 \frac{\arctan(x)}{x}\,dx\le \arctan(2) \int_1^2 \frac1x\,dx$$
A: \begin{align*}
\int_{1}^{2}\dfrac{\tan^{-1}x}{x}dx\leq\int_{1}^{2}\dfrac{\tan^{-1}2}{x}dx=(\tan^{-1}2)(\log 2)<1.
\end{align*}
A: Note that for $x\in [1,2],$  we have $$ 0.79 \le \tan^{-1}(1)<tan^{-1}(x)< tan^{-1}(2) \le  1.11$$ 
Thus $$\int_1^2 \frac{arctan(x)}{x} \,dx< 1.11 \int_1^2 \frac{1}{x} \,dx=1.11 ln(2)\le 0.77 <1.$$ 
A: Hint: It is shown geometrically in this answer, for $-\frac\pi2\lt x\lt\frac\pi2$,
$$
\frac{\tan(x)}x\gt1
$$
This means that
$$
\frac{\arctan(x)}x\lt1
$$
A: Alternative approach: $f(x)=\frac{\arctan x}{x}$ is a convex function on $[1,2]$, hence by the Hermite-Hadamard inequality
$$ \int_{1}^{2}\frac{\arctan x}{x}\,dx \leq \frac{1}{2}\left(\frac{\arctan 1}{1}+\frac{\arctan 2}{2}\right)=\frac{1}{2}\arctan(2+\sqrt{5})<\frac{\pi}{4}. $$
Second alternative approach: by the Shafer Fink inequality $\frac{\arctan x}{x}\leq \frac{\pi}{1+2\sqrt{1+x^2}}$, hence the integral is bounded by
$$ \int_{1}^{2}\frac{\pi}{1+2\sqrt{1+x^2}}\,dx \leq \frac{7}{10}.$$
