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Numerade Educator



Problem 15 Easy Difficulty

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

$ \displaystyle \int \frac{\coth \frac{1}{y}}{y^2}\ dy $


$$2 \sqrt{x} \arctan \sqrt{x}-\ln (1+x)+C$$


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Video Transcript

Okay, so this question wants us to evaluate this integral from our table. So again, this doesn't look like anything at our table. So we need to make a U sub. So it does. So based on the Y squared in the denominator and the one over when the argument maybe one over why would be a good choice for a use of and then do you becomes negative one over y squared d Y. But well, we don't have a negative one over y squared. We have a positive. So we need to multiply by negative one. So now we're good to go substituting and we get negative the integral of hyperbolic co tangent of you, do you? And that's a much simpler integral that we know how to calculate from our table. And our table just tells us that hyperbolic co tangent has an integral Ln of hyperbolic sign. So now all we got to do is plug back in for you and we're done. So we get negative, Ellen of the absolute value of hyperbolic sign both you, which is one over why plus c so again pretty straightforward. If you know what substitution to may