Solution
= Solution
Use the <substitution rule> $t=\tan x$, with $dt=(1+\tan^2x)\,dx$. The endpoints become $1$ and $\sqrt3$, so
$$
\int_1^{\sqrt3}\frac{dt}{(1+t)^2}=\left[-\frac1{1+t}\right]_1^{\sqrt3}=\frac12-\frac1{1+\sqrt3}=\boxed{1-\frac{\sqrt3}{2}}.
$$