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Evaluate the given definite integral.$$\int_{1}^{2} \frac{x}{1+x^{2}} d x$$

$$^{1} / 2(\ln 5-\ln 2)$$

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 6

The Definite Integral

Integrals

Campbell University

Baylor University

University of Michigan - Ann Arbor

Boston College

Lectures

05:53

In mathematics, an indefinite integral is an integral whose integrand is not known in terms of elementary functions. An indefinite integral is usually encountered when integrating functions that are not elementary functions themselves.

40:35

In mathematics, integration is one of the two main operations of calculus, with its inverse operation, differentiation, being the other. Given a function of a real variable (often called "the integrand"), an antiderivative is a function whose derivative is the given function. The area under a real-valued function of a real variable is the integral of the function, provided it is defined on a closed interval around a given point. It is a basic result of calculus that an antiderivative always exists, and is equal to the original function evaluated at the upper limit of integration.

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We're looking at the integral from 1 to 2 of X over one plus X squared DX, Um and what I would use is U substitution and let you it called the denominator and so d you will equal to X dx. Now, For me, it makes more sense to solve for DX if I'm going to substitute that in, and that's by dividing two X to the left side. So as I rewrite this in a rule, I like to leave X there. I can replace one plus x squared with you because that's what it equals. And then I can replace D X with 1/2 x d you because that's what d X equals. And I can visually see the excess eliminates. Um, I also have to change the bouncer because the top problem was in terms of X, and I need a change in terms of use. I go to right here and plug in one, and for X squared is one plus one is two now plugging two squared is four plus one is five and now I'm ready to evaluate the anti derivative. Now this one half is just a concept, so bring that in front the anti derivative of one over you is natural log of you and you don't need absolute value because where are bounds are going to be positive 2 to 5. And from here we're ready to plug in our bounce. We're looking at one half times natural to five May I will put it in the absolute value, Distribute practice minus natural geography, too. And what I would also do is I would actually because it's subtraction. We can combine them into a single log by divisions and natural log of five house. And this is how I would leave the answer. Um, if anyone is curious, you know, there are other correct answers. For instance, you can rewrite This is natural log of five halves to the one half power. We should be the same thing as natural log of the square root five has. But, you know, where do you stop? It's up to you

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