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Evaluate the integral.

$ \displaystyle \int \frac{y}{(y + 4)(2y - 1)}\ dy $

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$$\frac{4}{9} \ln |y+4|+\frac{1}{18} \ln |2 y-1|+C$$

Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 4

Integration of Rational Functions by Partial Fractions

Integration Techniques

Missouri State University

Campbell University

Boston College

Lectures

01:53

In mathematics, integration is one of the two main operations in calculus, with its inverse, differentiation, being the other. Given a function of a real variable, an antiderivative, integral, or integrand is the function's derivative, with respect to the variable of interest. The integrals of a function are the components of its antiderivative. The definite integral of a function from a to b is the area of the region in the xy-plane that lies between the graph of the function and the x-axis, above the x-axis, or below the x-axis. The indefinite integral of a function is an antiderivative of the function, and can be used to find the original function when given the derivative. The definite integral of a function is a single-valued function on a given interval. It can be computed by evaluating the definite integral of a function at every x in the domain of the function, then adding the results together.

27:53

In mathematics, a technique is a method or formula for solving a problem. Techniques are often used in mathematics, physics, economics, and computer science.

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Let's evaluate the following into girls. First thing we should do here is a partial fraction to composition. Looking at the denominator, we see, too linear terms, letting your factors here and they're distinct. So this is what the book would call case one. So in this case, we have a constant a over Y plus four and then another constant be over two or minus one. And then now we'LL just have to find out what a br and once we do, the right hand side will be easier to integrate. So the next thing we should do is it's good and multiply both sides a dissipation up here by the denominator y plus four times two one minus one. So Scott and more supply this to both sides on the left hand side. Since then, I'm the owner disappears, it cancels out, and then on the right hand side for the first term, the Y plus fours cancel out, and for the second term, the two y minus ones will cancel. And then let's just go ahead and rewrite this right hand side. It's good, in fact, are why there only to a plus B all in front of the Y and then we're left over with for be minus a So that's our constant. Or also we can rewrite this. Why, if we want is one why zero. So the term on the left hand side in front of the wise one, determine the right hand side in front of the wise to a plus B. So we must have to a plus B is one. Similarly, the constant on the left inside a zero. So the constant side on the right hand side also has to be zero. So we have four B minus a equal zero and let's go ahead and solve this two by two system For ambi many ways to do this, you could take the second equation and just move the dates of the other side. And then you could go out and plug in this value for a into the first equation. So doing so you get it. Ninety equals one. So be is one over mine and then is for B. So it has to be four times one over nine. So these are the values for Andy that will plug in up here and similarly for B. So we'LL plug those in the right hand side that will integrate. I'm running out of room here, So let me go to the next page. So are integral after the parcel. Fraction the composition. We have four over nine. That was R A. Why plus four. And then he was one of our nine. So that's the constant in the numerator. Over to my minus one. Now we have some minerals that we've seen before. These air thes will involve the log function natural algorithm. If this plus four is bothering you, you can go ahead and do it, you sub that ship's off resolved the problem and similarly for this term, if this too and the minus one are bothering you to a separate use up. So w w substitution two I minus one. In either case, after make After evaluating these integral sze, the first one becomes for over nine natural log absolute value of Y plus four. And for the second one, we'LL have one over eighteen. So this extra factor of two which you could see that it comes from here the u substitution d w equals two d y. And this is why we have to divide by two. And so we multiply the two and the night together to get eighteen. Then we have natural log to why minus one. And finally, don't forget that constant of integration C at the very end, and there's a final answer.

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