💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here!

Get the answer to your homework problem.

Try Numerade Free for 7 Days

Like

Report

Evaluate the integral.

$ \displaystyle \int_1^2 \frac{3x^2 + 6x + 2}{x^2 + 3x + 2}\ dx $

$$3+\ln (3 / 8)$$

Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 4

Integration of Rational Functions by Partial Fractions

Integration Techniques

Campbell University

University of Michigan - Ann Arbor

University of Nottingham

Idaho State University

Lectures

01:11

In mathematics, integratio…

06:55

In grammar, determiners ar…

06:37

12:29

02:40

06:46

08:47

08:10

01:48

00:42

Evaluate the indefinite in…

07:52

0:00

Evaluate the integral.…

Let's evaluate the definite in general from one to two we see that dancer grand. We have a polynomial divided by polynomial in the Palin oils have same degree for the numerator and denominator. They're both degree too. So we should do long division here polynomial division So three X where divided by X squared is three squatted Multiply out that three We get three x square plus nine x plus six and we subtract this whole thing So we should get minus three X and then to mine in six minus four Sense X square does not go into X. That means this down here is our remainder. So all this is that we can take this original into grand and right as the question was, three Then we have our remainder up top and then under bottom we have our original denominator, but let's actually go ahead and factor that So that'LL be X plus one and it's close to and the denominator It's so now we'LL do partial fraction the composition on this new fraction. So for analysis ignore the three on the left over here for a minute. So this is the partial fracture the composition will be a over X plus two. Please be over X plus one. So, my God, and set these equal to each other. This is what the book calls case one we have to linear factors in the Dominator and their distinct. So the next step would be to go ahead and multiply both sides of this equation down here by the denominator on the bottom left. So we get minus three X minus four on the left and on the right. After we multiply a times X plus one be times that plus two and we can go ahead. In fact, rather ex, we get a plus B and then for the constant term, we just get a plus two B Now, looking on the left hand side, we see that the coefficient in front of the X is a minus three. That means a plus B has to also equal negative three. And for the other equation or for the other coefficient we see here, the constant is minus four and the constant overhears a plus two b. So those also have to be equal, sir. So let's write these equations on the next page. We have a Plan B equals minus three. We also have a place to be equals minus four. So these were the equations that we Highland on the previous page. Now, let's go ahead and just solve this two by two system for Andy. So, for example, I could take this first equation over here, get a myself. Then we could go ahead and plug in this a value. And so the other question. So, by doing so, we had negative three must be equals minus four. Then go ahead and just solve that for B. Once we do that, put me back into this equation over here and saw Ferree. So have minus three. But then we're gonna have minus negative one. So it's a minus two. So now we finally have our coefficients for the partial fraction to composition. So at this point, with right out the interval Well, have been a girl want to too? We did the long division. We had a three before the partial fraction to composition. And then now we have plus a divided by X plus two and then be divided by X plus one. So this is where the and beer coming from and then Now we're ready to integrate. If this plus two and the plus one are bothering you here, you could go ahead and feel free to use U substitution. There. You could use that plus two. And over here you could use new equals. X plus one. Don't you come down here to this side. Bottom left after we integrate three X minus two Natural log X plus two minus natural log. Absolute value X plus one. And then our endpoints wants it too. Plugging into first six minus two natural log. A four minus natural. Log of three minus three minus two. Natural. Other three minus natural. Other, too. The next step here. So just simplify as much as we can. You see that you have six minus three. So you should get a three there. And that here it could also simplify the natural live of threes you could combine. See him minus one here and then plus two on the other side. So let's go to the next page three minus two. Ellen for plus Ellen three plus Alan too. Now this. There's many ways to go about this. Technically, we could stop here. Good. You might as well Simplify when you can. So, for example, if you wanted to use a lot of properties you khun Bright this too inside the exponents so would have minus ln foreswear which is sixteen plus And then we have Ellen of two times three. So, for example, we can write it this way and then we can use one of our lot properties. We have a lot of minus a log so we can write that is Lago a fraction and then finally just simplify that fraction and revenged. So here divide by two top and bottom and that's your final answer.

In mathematics, integration is one of the two main operations in calculus, w…

In grammar, determiners are a class of words that are used in front of nouns…

$ \displaystyle \int_0^1 \frac{2}{2x^2 + 3x +…

$ \displaystyle \int \frac{x^3 + 2x^2 + 3x - …

$ \displaystyle \int \frac{x^2}{x^6 + 3x^3 + …

$ \displaystyle \int_0^1 \frac{3x^2 + 1}{x^3 …

$ \displaystyle \int_{-1}^0 \frac{x^3 - 4x + …

$ \displaystyle \int \frac{x^3 + 6x - 2}{x^4 …

$ \displaystyle \int^2_{1} (4x^3 - 3x^2 + 2x)…

Evaluate the indefinite integral.

$ \displaystyle \int (x^2 + 1)(x^3…

$ \displaystyle \int_1^2 \frac{x^3 + 4x^2 + x…

Evaluate the integral.$\int \frac{\left(x^{2}+3\right) d x}{\left(x^{2}+…