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Use the Table of Integrals on Reference Pages 6-10 to evaluate the integral. $ \displaystyle \int \frac{\coth \frac{1}{y}}{y^2}\ dy $

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Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 6

Integration Using Tables and Computer Algebra Systems

Integration Techniques

Oregon State University

Harvey Mudd College

University of Nottingham

Idaho State University

Lectures

01:53

In mathematics, integration is one of the two main operations in calculus, with its inverse, differentiation, being the other. Given a function of a real variable, an antiderivative, integral, or integrand is the function's derivative, with respect to the variable of interest. The integrals of a function are the components of its antiderivative. The definite integral of a function from a to b is the area of the region in the xy-plane that lies between the graph of the function and the x-axis, above the x-axis, or below the x-axis. The indefinite integral of a function is an antiderivative of the function, and can be used to find the original function when given the derivative. The definite integral of a function is a single-valued function on a given interval. It can be computed by evaluating the definite integral of a function at every x in the domain of the function, then adding the results together.

27:53

In mathematics, a technique is a method or formula for solving a problem. Techniques are often used in mathematics, physics, economics, and computer science.

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Use the Table of Integrals…

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Use the Table of Integral…

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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

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