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Evaluate the integral.
$ \displaystyle \int \frac{dx}{1+ e^x} $
$\int \frac{d x}{1+e^{x}}=x-\ln \left|1+e^{x}\right|+C$
Calculus 2 / BC
Chapter 7
Techniques of Integration
Section 5
Strategy for Integration
Integration Techniques
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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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