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Make a substitution to express the integrand as a rational function and then evaluate the integral.

$ \displaystyle \int_0^1 \frac{1}{1 + \sqrt[3]{x}}\ dx $

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$-\frac{3}{2}+3 \ln (2)$

Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 4

Integration of Rational Functions by Partial Fractions

Integration Techniques

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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.

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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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Make a substitution to exp…

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Use a substitution to eval…

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Evaluate the integral.

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Find the Integral of \int …

let's use the substitution to rewrite this in a role as an integral, the rational function here. Many choices for you that may work here. Let's just try X to the wonder. Because of this extra, the wonder in the denominator take a differential on the side. So here is in the pot, a rule for derivatives. And I could rewrite this as one over three years squared. And the reason for that is because of this square, both sides. So you square and exit the two thirds of the same. And then I could go ahead and multiply the three squares to the other side. And I can replace the ex with this expression on the left. So let's go ahead and do that in our original problem. This now becomes integral, so we should also check the limits of integration. So to do that, we plug in the original bound zero and one into X and the U substitution. And when you do the U values that you get back or also zero and one in that same order, so X equals zero implies you equals zero, then X equals one implies you eagles won, then the X up top. That's just to re use. Where do you now? In the denominator? We just have one plus you Now, since the numerator has larger degree, let's go ahead and new polynomial division three squared over you. That's three. You go ahead and multiply that out with minus three U minus three. You divided by you minus three and then multiply that and we get a remainder of three. So this means that we can write our integral as the quotient, which was three. You minus three. Oh, and then we have our remainder and over you plus one, the original divisor. Now for the first term, she's a powerful for new girls. Same for the second. Here, this is natural log. Now let me make a comment about this plus one. If this plus one causing difficulty here, just go ahead and do it. Use up and coming back to our inner girl. We go to the end point zero and one supplied in the one first. You have three over, too. They have minus three plus three. Ellen, too. That's for plugging in one. And then when you plug in zero each of the three terms ghost zero and the third term goes to zero because natural Largo born zero. So we subtract off zero and then we just have three national log of two, and then we have minus three halves and there's a final answer.

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