`int_0^(1/sqrt(3)) (t^2 - 1)/(t^4 - 1) dt` Evaluate the integral

Expert Answers
Borys Shumyatskiy eNotes educator| Certified Educator

Hello!

`(t^2-1)/(t^4-1)=1/(t^2+1).`

Therefore indefinite integral is

`arctan(t)(+C).`

Substitute t from 0 to `1/sqrt(3)` and obtain

`arctan(1/sqrt(3))-arctan(0)=pi/6-0=pi/6 approx 0.524.`