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Evaluate the integral.

$ \displaystyle \int \frac{dx}{1+ e^x} $

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Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 5

Strategy for Integration

Integration Techniques

Harvey Mudd College

Baylor University

University of Michigan - Ann Arbor

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.

01:11

Evaluate the integral.…

01:16

02:51

Evaluate the integrals.

02:55

00:47

03:03

01:25

01:22

00:55

01:29

00:32

Evaluate the indefinite in…

0:00

evaluate Integral

01:06

02:09

Let's try to do a use up here. Let's take you to just be the denominator then do you? Is e the x DX? So here, let me just solve this for DX because in the original problem, we have the X by itself there up top. So do you Overeat of the X equals D X and then here solve this for either the X so we have d u over U minus one equals the X. So after the u sub, we can rewrite this so detox that just becomes D you u minus one and then we also need this on the bottom. But that's just you by our substitution. So that just gives us another factor down here, are you? And then we should use a partial fraction decomposition. Here's so let's look at that in a grant for a moment. Oh, so we have a over you be over u minus one. It's going to multiply both sides by that denominator on the left and then rewrite this by pulling out of you. So if you look at the coefficients here, you could find a two by two system. We have a plus. B equals zero because there's a you on the on the right. But there's no you on the left and similarly negative A equals one. So we get a equals negative one and plug this into this equation to get B equals one. Now that we have A and B, let's plug those in here and then we'll integrate this so we have a over you. So that's minus one over you and then one over U minus one. Now these we could integrate for the second integral right, If that plus one or that minus one is bothering you, feel free to do a substitution here. Another use of And also don't forget your constant of integration. See. So then, here we go back to our original use up u equals one plus e to the X, and then we just rewrite. This is negative, Ellen. Absolute value one plus e to the X plus natural log. Here I have u minus one. So that's just either the X plus e. Now, here we can drop the absolute value because one plus e to the X is always positive and similarly over here you can also drop the absolute value and then finally, here You can use the fact that the natural log and either the x r in verses of each other. So remember, when you compose a function in its inverse, they always cancel each other out and your left over with X. So, using that fact here, we're composing the exponential function and the log function so they'll cancel out there and we'll just have an ex left over. So plus X, plus our constancy, and that's our final answer.

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