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Evaluate the integral.

$ \displaystyle \int \frac{dx}{x (x^4 + 1)} $

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Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 5

Strategy for Integration

Integration Techniques

Missouri State University

Campbell University

Idaho State University

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.

00:55

Evaluate the indefinite in…

02:34

01:20

00:40

01:14

Evaluate the integral.…

01:43

for this in a girl. Let's go ahead and do a use up. You equals X players so that X equals you too. The one half And then here let's differentiate. So use that power rule and then we can rewrite X is you two the one half and then solve this equation here for D s So we have to do you over to use in the one half. Now let's rewrite this integral after the use of So here's RDX up here. Well, go ahead and replace that with do you over two square on you and then on the bottom we see that there's another exterior. So by our earlier work that'LL just be swear rule you. And then here we see exit of four plus one. So that's just X squared, squared plus one and then using the definition of you, that's just you. Square plus one. Now let's go ahead and combined. These use here that'LL just become also has pulled off the two we have won over you and then you square plus one. So before we try partial fracture decomposition, we should look at this quadratic here to see whether factors this one does not factor because the discriminated B squared minus four a c for this problem is negative, for which is less than zero. So we can write this in the form. And were you bu plus c overyou squared plus one and then do you and then of course, would have to go ahead and find and B and C So let me go on to the next page. Here we have one half and then is one. So we have one over x. So really here I should stick with you because that was just a variable, that word. And then I have minus and that you overyou squared plus one. So be is minus one cia zero. And then here let's go ahead and integrate this and that here we're getting the one half and then for this integral here you can go ahead and do it use up so d w over two equals you. Do you so that the one half is coming from this all here outside the integral. And then we get another one half here and then we have into rule and then we just have dw over and over You. So one half natural log, absolute value. You. And then here we see the one over four natural logged ofyou and it's an RC and no. Now let's go back and write everything in terms of X. So here w you squared plus one. But you was x square, so we could write that as X square square. So that's thanks to the four plus one. So we have one half Ellen and that by definition of you, this is just X square minus one over four natural lug. Excellent four plus one at our constant C at the and that's a final answer.

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