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Evaluate the integral.
$ \displaystyle \int \frac{\ln x}{x \sqrt{1 + (\ln x)^2}}\ dx $
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Calculus 2 / BC
Chapter 7
Techniques of Integration
Section 5
Strategy for Integration
Integration Techniques
Harvey Mudd College
Baylor University
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.
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Evaluate the integral.…
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Evaluate the integrals.
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Find the integral.$\in…
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Evaluate the indefinite in…
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Let's ease EU substitution here. Let's take you to be the natural log Then do you run and we see the one over X B c x here in the denominator and we see DX over there. So that corresponds to these two terms over here on the right. So after this use of we can rewrite this, we have a you appear less Ellen of X. And then do you then in the denominator, I have to rewrite this radical is one plus you square. So that's our new integral. And then here we could do another use a bliss to w equals one plus use. Where, then here, If you do dw and then divide by two that corresponds to this numerator here you do you so one half. Then we have dw up top and then square root of w on the bottom. So just use the power rule here. So one half then w to the one half power Then divide by one half at our constant of integration. See, cancel those one half and then replaced w and in terms of you So that becomes where one plus you square and then finally use the original U substitution u equals Ellen X to rewrite this. Not this square root, plus our constancy, and that's your final answer.
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