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Evaluate the integral.

$ \displaystyle \int^{2}_{1} \frac{(x - 1)^3}{x^2} \,dx $

$\int_{1}^{2} \frac{(x-1)^{3}}{x^{2}} d x=-2+\ln 8$

01:16

Frank L.

00:52

Amrita B.

Calculus 1 / AB

Chapter 5

Integrals

Section 4

Indefinite Integrals and the Net Change Theorem

Integration

Harvey Mudd College

Baylor University

University of Michigan - Ann Arbor

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right. The way I want to do this question is uh using the by normal serum, our pascal's triangle, which would be uh knowing that when you take something to the third power, Your coefficients will be 1, 3, 3, 1 or X squared. So to be X cubed an a minus X squared and then plus X and then minus one. If you don't believe me, you can foil that out three times. But there is a shortcut and that's using pascal's triangle. So now what I would do is divide each one of these pieces. So I'm still not doing the entire derivative executed by by exporters. Just X. These X squares would just cancel the next woman would become three over X. And then the last thing I'm going to write is X to the negative second power. There's a reason I'm writing all of these like this because now when I do the anti derivative where we add one to the exponent, multiply by the reciprocal of that exponent, we have an issue when we get right here. And that issue is that it needs to be natural log that's the anti derivative of one over X. But then this one's fine, you add one to the expo and multiply by the reciprocal. Uh and they were going from 1-2. So as we do all this shortcut and all of this work. Now we're ready to plug in our upper bounds, which is two squared is four half of that is to minus six plus three. Natural log of two. And then to to the 91st power is the same thing as 1/2. And we need to subtract off Plugging in one and for all those to be 1/2 minus three plus three natural log of one plus one. Now, as you're looking through all of that, I hope you notice at least the one half and the minus one half of later cancelled Natural Log of one is 0. Um So all we're left with on the right side is that negative three plus one? Which would be negative too. That when we're subtracting turns into plus tube so they have to Plus two is 4 -6 is -2. So if I've done all of my math, correct, this should be your correct answer. Um I'm happy with this answer. If you're curious how the answer key changed a little bit, you can use the power rule to change that to be an exponent. Uh and two cubed is eight. I'm not sure why anybody in their right mind would do this, but this is correct as well. Okay.

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