Integrals

[TWELVEPEANUTS11]

evaluate the integral

integral evaluated from 0 to 1/rad 3
(t^2 - 1)/ (t^4 - 1) dt

answer: pi/6

what i have done so far:
(t^2 - 1)/ (t^2 - 1)(t^2 + 1)
canceled t^2 - 1

left with 1/t^2 + 1

then integral of arctan t

but i do not know how to integrate that function, help would be much appreciated

 

[galactus]

You have \(\displaystyle \int\frac{1}{1+t^{2}}dt=tan^{-1}(t)\)

\(\displaystyle tan^{-1}(\frac{1}{\sqrt{3}})-tan^{-1}(0)=\frac{\pi}{6}\)

One usually does not derive this each time, but recognizes it and implements it accordingly.

 

[TWELVEPEANUTS11]

oh you're right. i kept thinking i had to integrate the arctan.
Thanks so much i see my error.