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Find the area of the region under the given curve from 1 to 2.

$ y = \dfrac{1}{x^3 + x} $

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Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 4

Integration of Rational Functions by Partial Fractions

Integration Techniques

Missouri State University

Oregon State University

Baylor University

University of Nottingham

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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Find the area of the regio…

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Find the area under the cu…

let's find the area of the region under the given curve from X equals one to two. So what they're asking for is theater girl from one to two of this function that'LL just be the area under the curve from wanted to And then we should do a partial freshen the composition here tto be able to evaluate this inaugural so we can rewrite that denominator. Now, this guy right here, this will not faster. If we just look at the Discriminate B squared minus four A. C will be negative. So this will not factor so we can write. This is ale Rex. That's what the author calls Case one and then we have B X plus C for the quadratic. And this is what the author calls Case three. It's got and multiply this equation on both sides by this term around here and then we can go ahead and combined term. So it's pulling X square. Here we have a plan B plus X. We just have CX and then a so comparing coefficients under left and right, we see a plus B must be zero c must be zero. They must be won. So we end up with It's right this out. April's one C equals zero, and by this over here be is negative one. Let's just go ahead and plug in these values for a B and C Hey into the right hand side over here and then we'LL evaluate this expression. Let's go to the next page. After plugging in a B and C, there's our inaugural. So for the second one, we can go ahead and he's a use up here, x squared, plus one and then do you over, too, so that may help you with this integral I won't actually use the use of here. So the first annual girl Natural log and then for the second minus one half you could see where the one half is coming from. Natural log X squared, plus one wanted to, and then just go ahead and plug in the end. Points in Simplify Natural law, too. One half Ellen five plus one half Ellen, too. And then we have three half combining those natural log of twos, and there's a final answer

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