Evaluate integral from 0 to sqrt3 of (t^2-1)/(t^4-1) dt.

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

(1) First factor the denominator as `(t^2+1)(t^2-1)` . After cancelling we have `int_0^sqrt(3)(dt)/(t^2+1)`

(2) Recognize this as the arctan: `int(du)/(u^2+a^2)=1/a arctan( (u)/(a))+C`

(3) So `int_0^sqrt(3)(dt)/(t^2+1)=(1/1)arctan(t/1)|_0^sqrt(3)`

or `arctan (t)|_0^sqrt(3)=arctan (sqrt(3))-arctan (0)=pi/3-0=pi/3`

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

(1) First factor the denominator as `(t^2+1)(t^2-1)` . After cancelling we have `int_0^sqrt(3)(dt)/(t^2+1)`

(2) Recognize this as the arctan: `int(du)/(u^2+a^2)=1/a arctan( (u)/(a))+C`

(3) So `int_0^sqrt(3)(dt)/(t^2+1)=(1/1)arctan(t/1)|_0^sqrt(3)`

or `arctan (t)|_0^sqrt(3)=arctan (sqrt(3))-arctan (0)=pi/3-0=pi/3`

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