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

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

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Anna Marie V.

Campbell University

Heather Z.

Oregon State University

Caleb E.

Baylor University

Boston College

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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.

Topics

Integrals

Integration

Anna Marie V.

Campbell University

Heather Z.

Oregon State University

Caleb E.

Baylor University

Boston College

Lectures

Join Bootcamp