Answer
$$\int^{1/\sqrt3}_{0}\frac{t^2-1}{t^4-1}dt=\frac{\pi}{6}$$
Work Step by Step
$$A=\int^{1/\sqrt3}_{0}\frac{t^2-1}{t^4-1}dt$$ $$A=\int^{1/\sqrt3}_{0}\frac{t^2-1}{(t^2-1)(t^2+1)}dt$$ $$A=\int^{1/\sqrt3}_{0}\frac{1}{t^2+1}dt$$ According to Table 1, we have $$\int\frac{1}{x^2+1}dx=\tan^{-1}x+C$$ Therefore, $$A=(\tan^{-1}x)\Bigg]^{1/\sqrt 3}_{0}$$ $$A=\tan^{-1}(\frac{1}{\sqrt 3})-\tan^{-1}0$$ We know that $\tan^{-1}(\frac{1}{\sqrt 3})=\frac{\pi}{6}$.
For $\tan^{-1}0$, we see that as $\tan x=0$, $\sin x$ must also equal $0$, since $\tan x=\frac{\sin x}{\cos x}$. And we have $\sin0 =0$, so $\tan^{-1}0=0$. $$A=\frac{\pi}{6}-0$$ $$A=\frac{\pi}{6}$$
