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Evaluate the integral.

$ \displaystyle \int \frac{x^4}{x - 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

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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Evaluate the indefinite in…

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Evaluate the integral.…

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01:58

Evaluate the definite inte…

Let's evaluate the integral of X to the fourth over X minus one. Here, the first thing we noticed is that the numerator has larger than read. So let's go out in these long division X minus one going into X to the fourth power we have executed. We'Ll supply that subtract left over with X cubed so we can add X squared up here we had X cubed minus X squared That's over with X squared so we could come back up here at an ex and then we'll have X squared minus x subtract left over with X And then we could come back here once more and then add that one in there and then we're X minus one Subtract that off and we had a one left over X Does not go into one. So this is our remainder. So we could come and write this as well. We just did the division our quotient was X cubed Ex player accident one and then we have our remainder divided by the question the original divisor X minus one and this is a much easier in the world. In the original, we can use the power rule for the first four terms for the last one, and may or may not help you to use the use up here. If you need to use because of the minus one, go ahead and do the U equals X minus one. So using the power rule and then the natural log, we have extra the fourth over. Four. Execute over three X squared over two X and then plus natural law. Don't forget that. Absolute value X minus one inside and closer constancy of integration and there's our answer.

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