∫ 0 1 3 t 2 − 1 t 4 − 1 d t = ∫ 0 1 3 ( t 2 − 1 ) ( t 2 − 1 ) ( t 2 + 1 ) d t = ∫ 0 1 3 1 ( t 2 + 1 ) d t = t a n − 1 | 0 1 3 {\displaystyle {\begin{aligned}\int _{0}^{\frac {1}{\sqrt {3}}}{\frac {t^{2}-1}{t^{4}-1}}dt=\int _{0}^{\frac {1}{\sqrt {3}}}{\frac {(t^{2}-1)}{(t^{2}-1)(t^{2}+1)}}dt=\int _{0}^{\frac {1}{\sqrt {3}}}{\frac {1}{(t^{2}+1)}}dt\\[2ex]&=tan^{-1}{\bigg |}_{0}^{\frac {1}{\sqrt {3}}}\end{aligned}}}
![{\displaystyle {\begin{aligned}\int _{0}^{\frac {1}{\sqrt {3}}}{\frac {t^{2}-1}{t^{4}-1}}dt=\int _{0}^{\frac {1}{\sqrt {3}}}{\frac {(t^{2}-1)}{(t^{2}-1)(t^{2}+1)}}dt=\int _{0}^{\frac {1}{\sqrt {3}}}{\frac {1}{(t^{2}+1)}}dt\\[2ex]&=tan^{-1}{\bigg |}_{0}^{\frac {1}{\sqrt {3}}}\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/23256191d9a68f1123815a0d78389bd930701c3f)