Answer
$$
\int_{1 / \sqrt{3}}^{\sqrt{3}} \frac{8}{1+x^{2}} d x =\frac{4 \pi}{3}
$$
Work Step by Step
$$
\begin{aligned}
\int_{1 / \sqrt{3}}^{\sqrt{3}} \frac{8}{1+x^{2}} d x &=8\int_{1 / \sqrt{3}}^{\sqrt{3}} \frac{d x}{1+x^{2}} \\
&=[8 \arctan x]_{1 / \sqrt{3}}^{\sqrt{3}} \\
&=8\left(\arctan (\sqrt{3}) -\arctan (1 / \sqrt{3}) \right) \\
&=8\left(\frac{\pi}{3}-\frac{\pi}{6}\right) \\
&=8\left(\frac{\pi}{6}\right) \\
&=\frac{4 \pi}{3}
\end{aligned}
$$
