Evaluate the integral or state that it diverges. ∫2∞1x(lnx)4dx\int_2^{\infty}\frac{1}{x(lnx)^4}dx∫2∞x(lnx)41dx

1 3 ( ln ⁡ 2 ) 3 \frac{1}{3\left(\ln2\right)^3} 3 ( l n 2 ) 3 1 ; converges.

Evaluate the integral or state that it diverges. ∫ 2 ∞ 1 x ( l n x ) 4 d x \int_2^{\infty}\frac{1}{x(lnx)^4}dx ∫ 2 ∞ x ( l n x ) 4 1 d x

this problem, we are asked to evaluate the following integral and state whether or not it diverges. So here I have the integral that goes from two to infinity. Now since I can see infinity shows up in one of the bounds here, I know that I'm dealing with an improper integral. How do we deal with these? Well, when it comes to improper integrals, what I need to do is rewrite this as a limit as t approaches this upper bound of infinity. So that way my integral is gonna go from two to now this upper bound of t. Now remember, limits, we can have these approaching infinity. Limits can go to infinity. So in this case, we can put this t as an upper bound, and that's gonna be the integral of one over x times the natural log of x to the fourth power integrated with respect to x. So now we have our integral rewritten. Now what do I do from here? Well, at this point, I need to figure out how I can evaluate this integral. And since I can see we have this rational function we're dealing with, I'm first going to see if there's somewhere I can do a variable substitution. Now what I'm going to do is try a variable substitution on this natural log of x down here. So what is that going to look like? Well if I go ahead and set that natural log of x equal to u, well we're going to keep this limit continually on the outside, but that's going to give me the integral from two to t. We're going to have one over x, and then we're setting this whole piece equal to u right here, the natural log of x, so that means we're going to have u to the fourth power integrated all with respect to x. So this is our function. So now here we're saying that u is equal to the natural log of x. Now at the moment, I cannot integrate this because we still have x's that we're dealing with, and we have u's. We don't want two variables to be in our function when we're trying to integrate. But what I can do is take the derivative of u with respect to x, and we should know that the derivative of the natural log of x is just gonna be one over x with respect to x. Notice something that happens here. We get a one over x d x right here. And notice we have a one over x dx right there, so we can just replace that whole thing with du. So going ahead and rewriting this integral all out, we're going to keep this limit, and then that's going to be from two to t, that's going to be one over u to the fourth du. And now this is something I can integrate. Now my first step would be to take this u to the fourth power and rewrite it as u to the negative fourth power, since I can see it's in the denominator with one in the numerator, so we're gonna have two to t, and that's gonna be u to the negative fourth power du. Now what do I do from here? Well from here what I need to go ahead and do is the power rule on this u. So doing that will add one to the exponent and then divide by that new exponent. So that's going to give me that this whole thing is equal to the limit as t approaches infinity for the integral from two to t. And then in this case, we have u to the negative fourth plus one over negative four plus one. Once we go ahead and apply that power rule, we've done the integral step. So that's just going to give me u to the negative three over negative three bounded from two to t. Now at this point, I'm going to rewrite and simplify what I have inside these brackets. So go ahead and keep this limit on there, and rewriting this, I could write this as one over and it's gonna be negative three u cubed, just bringing that u to the negative three power in the denominator. And this whole thing is bounded from two all the way up to t. Now I can't deal with my bounds yet because notice that we have these pieces when we were originally in terms of x. We had x originally, which we switched to u, so I need to switch this back to x. Well, I can do that by recognizing u as just the natural log of x. So going down here, I can write this. So so we're going to keep this limit on the outside here, that's going to be negative one over three times the natural log of x cubed, and this whole thing will go from two all the way up to t. And what I'm first gonna do is plug in my t. So plugging in t, and for x in this function, we get negative one over three times the natural log of t cubed. Then what I'm going to do is subtract off my lower bound plugged in. So subtracting this off from this, those negative signs will cancel, so we get plus one over, in this case, it's going to be three times the natural log of two cubed. So this is what we end up with. Now what I need to do from here is apply my limit. What's gonna happen as this t approaches infinity? Well, as this t approaches infinity, notice we would end up with this number in the denominator that's getting bigger and bigger and bigger. This whole thing is being cubed, so what's gonna happen is this whole denominator is going to grow and insane insanely explode. Because this whole thing is just gonna get infinitely big, it's going to overtake the numerator setting this whole thing to zero. So once we apply this limit here, all we're going to end up with is one over three times the natural log of two cubed. And this right here would be the solution to our improper integral. So that is how you can solve this problem where we had to evaluate this improper integral up here. And because we went ahead and got a result that showed up, a finite number or a finite value, we could say that this whole thing converges. So So we would say this converges to this value one over three times the natural log of two cubed. So that is how you can solve this problem. Hope you found this video helpful. Let me know if you have any questions, and let's move on.

Evaluate the integral or state that it diverges. ∫ 2 ∞ 1 x ( l n ⁡ x ) 4 d x \int_2^{\infty}\frac{1}{x(ln⁡x)^4}dx ∫ 2 ∞ x ( l n ⁡ x ) 4 1 d x

1 3 ( ln ⁡ 2 ) 3 \frac{1}{3\left(\ln2\right)^3} 3 ( l n 2 ) 3 1 ; converges.

1 5 ( ln ⁡ 2 ) 5 \frac{1}{5\left(\ln2\right)^5} 5 ( l n 2 ) 5 1 ; converges.

11. Techniques of Integration

Improper Integrals

Business Calculus

11. Techniques of Integration

Improper Integrals

Struggling with Business Calculus?

Join thousands of students who trust us to help them ace their exams!Watch the first video

Multiple Choice

Evaluate the integral or state that it diverges.
$\int_2^{\infty}\frac{1}{x(lnx)^4}dx$

The Integral diverges.

$2$ ; converges.

$\frac{1}{5\left(\ln2\right)^5}$ ; converges.

$\frac{1}{3\left(\ln2\right)^3}$ ; converges.

Verified step by step guidance

Step 1: Recognize that the given integral is an improper integral because the upper limit of integration is infinity. This means we need to evaluate the integral as a limit: $ \int_2^{\infty} \frac{1}{x(\ln x)^4} dx = \lim_{b \to \infty} \int_2^b \frac{1}{x(\ln x)^4} dx $.

Step 2: Perform a substitution to simplify the integral. Let $ u = \ln x $, which implies $ du = \frac{1}{x} dx $. When $ x = 2 $, $ u = \ln 2 $, and as $ x \to \infty $, $ u \to \infty $. The integral becomes $ \int_{\ln 2}^{\infty} \frac{1}{u^4} du $.

Step 3: Simplify the new integral. The integral $ \int \frac{1}{u^4} du $ can be rewritten as $ \int u^{-4} du $. Use the power rule for integration: $ \int u^n du = \frac{u^{n+1}}{n+1} + C $, where $ n \neq -1 $. Here, $ n = -4 $, so the integral becomes $ \frac{u^{-3}}{-3} = -\frac{1}{3u^3} $.

Step 4: Apply the limits of integration. Substitute the limits $ u = \ln 2 $ and $ u \to \infty $ into the result of the integral. The expression becomes $ \lim_{b \to \infty} \left[ -\frac{1}{3u^3} \right]_{\ln 2}^b = \lim_{b \to \infty} \left( -\frac{1}{3b^3} + \frac{1}{3(\ln 2)^3} \right) $.

Step 5: Evaluate the limit. As $ b \to \infty $, the term $ -\frac{1}{3b^3} $ approaches 0. Thus, the result of the integral is $ \frac{1}{3(\ln 2)^3} $. Since this is a finite value, the integral converges.

11:11m

Watch next

Master Improper Integrals: Infinite Intervals with a bite sized video explanation from Patrick

Start learning

Improper Integrals practice set

Problem sets built by lead tutors

Expert video explanations

Go to Practice