Calculus (3rd Edition)

$$\ln \frac{4}{3}.$$
Let $u= \tan^{-1}x$, then $du =\frac{dx}{x^2+1}$ and when $x:1 \to \sqrt 3$ then $u: \frac{\pi}{4}\to \frac{\pi}{3}$. Now, we have $$\int_1^{\sqrt 3} \frac{ dx}{\tan^{-1}x(x^2+1)}=\int_\frac{\pi}{4}^\frac{\pi}{3} \frac{du}{u}\\ =\ln u|_\frac{\pi}{4}^\frac{\pi}{3}=\ln \frac{\pi}{3}-\ln \frac{\pi}{4}=\ln \frac{4}{3}.$$