\int_1^{+\infty} \frac{\arctan x}{x^2 \sqrt{x^2-1}}\, dx = \int_1^{+\infty} \arctan x \, \,d\left(\frac{\sqrt{x^2-1}}{x}\right)=\Bigl[\arctan x \frac{\sqrt{x^2-1}}{x} \Bigr]_{x=1}^{x\to +\infty}-\int_1^{+\infty} \frac{\sqrt{x^2-1}}{x} \frac{1}{x^2+1}\, dx=\frac{\pi}{2}-\int_1^{+\infty} \frac{\sqrt{x^2-1}}{x} \frac{1}{x^2+1}\, dx