∫ 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 = arctan ( t ) | 0 1 3 = arctan ( 1 3 ) − arctan ( 0 ) = π 6 − 0 = π 6 {\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]&=\arctan {(t)}{\bigg |}_{0}^{\frac {1}{\sqrt {3}}}\\[2ex]&=\arctan \left({\frac {1}{\sqrt {3}}}\right)-\arctan(0)\\[2ex]&={\frac {\pi }{6}}-0\\[2ex]&={\frac {\pi }{6}}\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]&=\arctan {(t)}{\bigg |}_{0}^{\frac {1}{\sqrt {3}}}\\[2ex]&=\arctan \left({\frac {1}{\sqrt {3}}}\right)-\arctan(0)\\[2ex]&={\frac {\pi }{6}}-0\\[2ex]&={\frac {\pi }{6}}\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/094d08285abbfc9a04df77ce69f091702bf79da4)