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Find the general indefinite integral.

$ \displaystyle \int \biggl( \frac{1 + r}{r} \biggr)^2 \, dr $

$-\frac{1}{r}+2 \ln r+r+C$

01:03

Frank L.

00:49

Amrita B.

Calculus 1 / AB

Chapter 5

Integrals

Section 4

Indefinite Integrals and the Net Change Theorem

Integration

Oregon State University

University of Michigan - Ann Arbor

Idaho State University

Boston College

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the way I would do this. And enroll is just think about what it means to be uh something squared and you can just square each piece. So I'm just rewriting the problem the way it was written first like this and then off to the side. I'm thinking about doing some scratch work of what does it actually mean to be one plus R squared? What means multiplied by itself? So as you distribute each piece in there, I think it's pretty straightforward that you get 11 hour plus one hours to our plus R. Squared. And then the other thing is instead of just dividing everything by R squared in divide each piece by R squared. Um Because think about if you're dividing what you need the same denominator in order to combine them together. So I'm trying to just simplify this so that the inner role makes a lot more sense. Um This would be to over our, I like to leave it like that and you'll see why in a second. Plus one is when we go to take the derivative. So those are just some scratch work. Um You add one to your exponents and then multiply by the reciprocal through new experiment. Now this one if you were to leave this as ardent and then the first and add one, you would get to the zero and then you'd be divided by zero is undefined. So it makes sense then that there must be a different function that gives us that derivative and that answers to natural log about. And you can double check that's right. If you did a derivative of two natural log of our, you get that answer. The directive of that anti directive that is just are and don't forget about plus C. Um Now I would rewrite this. Just kind of out of habit like this and don't forget you need that plus C. Because uh the derivative of a constant zero and they're not going to write plus zero up here. So this is a good answer.

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