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Factor $ x^4 + 1 $ as a difference of squares by first adding and subtracting the same quantity. Use this factorization to evaluate $ \displaystyle \int \frac{1}{(x^4 + 1)}\ dx $.

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Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 4

Integration of Rational Functions by Partial Fractions

Integration Techniques

Campbell University

Oregon State University

Harvey Mudd College

University of Michigan - Ann Arbor

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.

16:02

Factor $x^{4}+1$ as a diff…

00:56

Evaluate the integral.…

03:48

point)Evaluate the int…

0:00

Evaluate the integral.

Let's go ahead in fact, their ex to the fourth plus one. So here we can use the following. Let's go ahead and add in a two X square there and then we'LL make up for it by subtracting it at the end. And then we can write this as X Square plus one square and then minus roots. Who Times X and that's all square too. And then here on this latest expression, we could do difference of squares. There's the first and then here's the second. So now we can rewrite this just the instagram, not the integral itself. So we can go out a cheque that these guys are irreducible B squared minus four a c For both of these, quadratic is negative. So these do not fatter over the real numbers. So, using what the author calls case three, our partial fraction decomposition should look like the following. And then CX plus de X squared wound to X plus one x there plus one. It was sloppy over here. Sorry about that, she x Okay, Now let's go to multiply both sides of this equation by the denominator on the left. So we have one on the left X plus be. And then here you should have had a one there. Sorry about that. And then we have X square plus one to X plus one and similarly for the CX plus the X squared minus room to X plus one. Now, let's go ahead and combine like terms here. Our first lesson's going and expand the school thing. After we expand and simplified the right hand side, we should give a plus e. So I did skip a short step there. You would have to expand the right hand side first. And now I'm just pulling out the X cube that squared and so on. All right. That should not be there next time. So then you pull out the ex term a route B C minus room, soon, dear. And then now, at this point, you would compare match up the coefficients to see that a plus c is zero the second term zero, the third term zero and then B plus he is one. This is a four by four system for us to solve here for a, B, C and D. So we do have a plus. C equals zero. We can go ahead and plug in this fact. We have a and we have a seat here. So we end up with a room to be minus room to D equals zero. This implies being equals. If we plug this factor into this equation here we get B equals three equals a half. And now, similarly, we could go ahead and soft fur A A and C because we have one equation appear a plus, equal zero and then plug in our values for be Indians to this equation. Here you get to a you two seem equals minus one. So solve this two by two system over here on the left for and see. So here you could go ahead and multiply both sides of this equation by route, too. And then go ahead and add those together. And then sulphur, eh? And then you get C equals one over to room two. So now we have our values for a B C. And let's just go ahead and plug these into the animal for Are these air? Now? We're plugging into our partial fraction to composition. So in a girl over here, or denominator which will deal within a second will have to complete the square on the denominator. And then let me just break this inaugural into two. And then also over here, same denominator. Well, not the same. This one has a plus. And the ex is not inside the round court system too. Now, we could go ahead and take this, that I'm Nader completely square. So let's replace both of the nominators with this. And then they're after this point, we could go out and do you some. So D'You equals the X and we could also right That's equals you, plus one of the route to So it's going and rewrite the numerator term here. So working on this interval first negative one over to two. Then you close one over and then root, too. Using this equation here for X plus, we'LL have So this is negative one over two you plus one fourth. So now let's go ahead and rewrite this first Integral all in terms of you saw the next page for that plus one fourth and then on the bottom We have used where plus one half So let's go ahead and simplify this Pull out the constant you these flare plus a half on. Then let me go ahead and pull up that one force. So just playing this animal up inside, too. Damage is right. This is one half again then, do you? Well, do you here too now for this? Feel free to do you some from the center bro here and then for the second. And overall, we should do it tricks up unless you memorize something from the table. So go ahead and use thes substitution to evaluate these generals. And then I'll just for a few square. No absolute value here because he's squared. Plus a half is positive. And then after the truths of or if you've memorized the table, we have this and then we can go ahead and plug everything back in terms of X and also here, I should have had a additional one half from the user. Sorry about that. That's where the force coming from two times two. So just replacing you in terms of X and then here is well, so now that that takes care of the first integral from the previous, we still have the second integral. So for the second interval, you would just go ahead and use the substitution X plus one over radical to instead and then for the other in the grill of the second one that we didn't evaluate. So both of these intervals are very similar. We went over the first one in full detail, and the second one is too similar to write out a full detail again here. So this would be the second interval and then adding the results. Together we have the previous integral from the last page. So this was our first integral, that we weren't out in full detail. And then now we add the answer from this page. Then we have two x plus y, and then now finally will go ahead at a constant of immigration, see? And that's the final answer.

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