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Problem 16 Easy Difficulty

Use the Table of Integrals on Reference Pages 6-10 to evaluate the integral.

$ \displaystyle \int \frac{e^{3t}}{\sqrt{e^{2t} - 1}}\ dt $


$\frac{e^{t}}{2} \sqrt{e^{2 t}-1}+\frac{1}{2} \ln \left|e^{t}+\sqrt{e^{2 t}-1}\right|+C$


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Video Transcript

Okay, This question wants us to evaluate this integral using our table. So to do that, we need to get this into a form that our table can recognize. So we see we have a square root in the bottom and usually with square roots, we like a you squared in there somewhere. So let's call each of the two t you squared. So if we square root both sides, that means that you is equal to eat a t. So d'you is equal to either the t d t. Then we can rewrite this integral so we can see what's going on here. So we get the integral of each of the two tee times e to the t d t. We're just splitting up the top here times the square root of eating the two tea minus one. And we know that each of the T d t is, do you? And then the each of the two teas that are left just become you squared. So we have a use squared times a d. U over square roots of you squared minus one. And now this is in the form that our table recognizes. So we see that this integral evaluates to the following expression. This should be a minus. Sign here, though. Sorry. And now we have to do is plug in you and plug in a So a squared is one. And to you is e to the t So plugging that in we get eating that sea over two times the square root of each of the two tea, minus one plus 1/2 times the natural log of either that C plus square root of B to the two t minus one plus c and that's our final answer.

University of Michigan - Ann Arbor
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