In Exercises 35–68, use integration, the Direct Comparison Tes...

Question

In Exercises 35–68, use integration, the Direct Comparison Test, or the Limit Comparison Test to test the integrals for convergence. If more than one method applies, use whatever method you prefer.

∫ from 1 to 2 of (dx / (x ln x))

Accepted Answer

Welcome back everyone. Determine whether the improper integral from 0 to 1/2 of d x divided by x ln squared of x converges or diverges. A converges and b diverges. For this problem, let's notice that the function that we're integrating is undefined at x equals 0, right? Division by 0 is undefined, and ln of 0 is undefined as well. So what we're going to do is rewrite the integral as a limit. That would be a limit of the integral from B to 1/2. As B approaches. 0, right? And specifically we're going to show that B approaches 0 from the right. Considering the natural log of x function and we're basically integrating d x. Divided by X. L n squared of x In order to integrate, we're going to use use substitution. Let's suppose that U is a line of x. And then the derivative of u with respect to x is going to be 1 divided by x. So now DX divided by X is going to be DU, right? We can now multiply both sides by dx to show that DU is equal to dx divided by x. Now let's change the limits of integration so the lower 1, u1 is going to be. Alan of B. And the upper 1, U2 is going to be. Ellen of one half. Which can also be written as negative of 2 using the properties of logarithms. So now let's rewrite the integral. That would be limit as b approaches 0 from the right. Of the integral. From Helen of B. Up to negative alarm of 2 of. DX divided by x becomes DU. And we're dividing by ln 2 of X, which is U 2. So what we're going to do is apply the power rule to integrate and we can just rewrite the integrant SU to the power of -2DU. Now let's integrate. We get limit as B approaches 0 from the right. Using the power rule, we get to use the power of -1 divided by -1. And we can factor out the constant, which is -1, right? We can also rewrite you to the power of -1 S1 divided by u. And evaluate the result from Llan of B up to negative LN2. Let's go ahead and do that. We get negative limit. As B approaches 0 from the right. Off 1 divided by Alan of 2 minus. 1 divided by ln of b. And now let's calculate the result. Distributing the negative sign we get. 1 divided by LN of 2. Right? Minus negative again we get positive 1 divided by ln of. 0 from the right. So, the last term is undefined, but we can consider the limit because that's what we have, right? So Eanov 0 from the right is negative infinity. We can show that. En of b as b approaches 0 from the right is negative infinity. And a constant divided by. An infinitely large number is going to give us a limit of 0, so that final term gives us a limit of 0. And therefore the final answer is 1 divided by eon of 2. Therefore, the integral converges since we got a finite value, right? So the correct answer is option A. Thank you for watching.

In Exercises 35–68, use integration, the Direct Comparison Test, or the Limit Comparison Test to test the integrals for convergence. If more than one method applies, use whatever method you prefer.∫ from 1 to 2 of (dx / (x ln x))

Watch next

In Exercises 35–68, use integration, the Direct Comparison Test, or the Limit Comparison Test to test the integrals for convergence. If more than one method applies, use whatever method you prefer.
∫ from 1 to 2 of (dx / (x ln x))