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Numerade Educator

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Problem 43 Hard Difficulty

Evaluate the integral.

$ \displaystyle \int^{1/\sqrt{3}}_{0} \frac{t^2 - 1}{t^4 - 1} \,dt $

Answer

$$\frac{\pi}{6}$$

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Video Transcript

So here we want to start by factoring now this denominator we have. So it's only right that 1/3. So we have a numerator t squared minus one. So we can write T the 4th -1 as so t squared last one multiplied by t squared plus one. So now we can cancel out the numerator in his first term of the dominator. See we'll get one over T squared plus one. And now we can refer to the Table of indefinite Girls. Which will tell us that this integral will be inverse tangent or arc Itandje in of t Upper bound 1/3. Lower bound of zero. So let's evaluate this. Our tangents foreign over route three minus or a tangent of zero. Arc tangent 1/3 Will be Pi over six. Whatever route through. Excuse me, Might start changing of zero this 0. So our answer is Pi over six.