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Use the Table of Integrals on Reference Pages 6-10 to evaluate the integral. $ \displaystyle \int \frac{x^4\ dx}{\sqrt{x^{10} - 2}} $

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Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 6

Integration Using Tables and Computer Algebra Systems

Integration Techniques

Campbell University

Harvey Mudd College

University of Michigan - Ann Arbor

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.

03:06

Use the Table of Integrals…

03:50

Use the Table of Integral…

03:15

02:42

05:43

02:48

07:55

04:20

02:16

Okay, so this question wants us to find another anti driven. So to do so, we're gonna want to put this into the form of our table into gross. So we usually like to see some sort of a you squared and our denominator So we want you squared to be extra the 10th and that implies that you has to be X to the fifth. So d'you is five x to the fourth d x, but we only have one x to the fourth. So that means that d you will for five is extra the fourth DX. So now we can substitute So we have the integral of d U Over pulling the 1/5 in front the square roots of you squared minus two. And now if we look, here's what are integral Table tells us so In this case, a squared equals two. So that means that our anti derivative is 1/5 from our general factor times Ln of you plus square root, you squared minus hey squared which is two plus c. And now we can simplify this as 1/5 time's the natural log of Well, now we just gotta back substitute for you, which we said was X to the fifth X to the fifth plus square roots of you squared, which is X to the 10th minus a squared, which is two plus c. So a relatively simple looking anti derivative for a complicated in secret.

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