Answer
$$
\int_{0}^{1 / \sqrt{3}} \frac{t^{2}-1}{t^{4}-1} d t =\frac{\pi}{6}
$$
Work Step by Step
$$
\begin{aligned}
\int_{0}^{1 / \sqrt{3}} \frac{t^{2}-1}{t^{4}-1} d t &=\int_{0}^{1 / \sqrt{3}} \frac{t^{2}-1}{\left(t^{2}+1\right)\left(t^{2}-1\right)} d t\\
&=\int_{0}^{1 / \sqrt{3}} \frac{1}{t^{2}+1} d t \\
&=[\arctan t]_{0}^{1 / \sqrt{3}}=\arctan (1 / \sqrt{3})-\arctan 0 \\
&=\frac{\pi}{6}-0 \\
&=\frac{\pi}{6}
\end{aligned}$$
