Hello and welcome to my channel. I am ENA Agustin, please if you are new here, kindly subscribe to my channel. And if you have been watching my videos and not just subscribe, please kindly subscribe for us to build the mathematics community. I welcome you to my series on integration by pass. So in this lesson, you will solve some examples to improve your understanding on integration by pass. So join me as you go to nuts, remember to like share subscribe and leave your comments. Now, the first question that.

You're going to take is the integral of x to the power, 4 times, natural log of x all squared DX. Remember, this is less than 3 on this series. So check the description for lesson one. And then lesson two here, you pull down the integration by path formula. Integral u.

D, v is equal to u v, minus the integral then u, then choose u. And given by using the acronym is. So here this is investigative function. Logarithmic functions, algebraic technology and exponential function. So if we look to we don't have any.

Invest by half logarithmic function. So we take u to equal to natural log of x r squared. And it has left like this.

So we take our division to be equal to x to the power, 4 DX. That is algebraic function. Now we have to differentiate this to find our u.

Then we have to integrate this to find our infinity. So let's do this. If we integrate given x to the power, four DX, they would have been to equal to x to the power, five and five. They have now finished. So there's, no need bringing the integral constant. And.

Here if we let c to equal to natural log of x, then it means that u is equal to t squared the derivative of u with respect to t is equal to 2t, which is equal to 2 times natural log of x and the derivative of t with respect to x is equal to 1 divided by x. If we define the natural log of x, equal 1, divided by x, uh. If you want videos on derivative of natural logarithmic, for instance, check the description for that video.

So here the derivative of u with respect to x will be the derivative of t with. Respect to x times the derivative of u with respect to t, and that will give us 1 divided by x times 2 times natural log of x. And that is 2 times natural log of x divided by x. Wow. So here we have our d u on the x to be equal to 2 times natural log of x divided by x. Now let's make the u distance by multiplying 2 by DX.

So this implies that DX sorry u is equal to 2 times natural log of x divided by x DX. So now we are going to go so here I want to have the integral you define. This is our u. And this.

Is our division, so we have x to the power 4 natural logs of x or squared, or let me write it happen before the same thing it's equal to u v. This is my u. And this is my gain. So I have x to the power 5, divided by 5 times, natural log of x, minus the integral then x to the power 5, divided by 5 times Du that is 2 times natural log of x divided by x DX. So that this will cancel this. We have x to the power 5 and can factorize 1 over 5 out. So we are going to get this to be equal to x to the power 5, divided. By 5 times natural log of x to the power 2, minus 1 divided by okay, 2 is here so 2, divided by 5, rather so 2, divided by 5, the integral x to the power 4 times natural log of x DX.

Now we are going to take this one through and integrate using integral by paths again. So let's, take this portion. So for the integral x to the power, four times natural log of x DX. We take u to equal to natural log of x and divide equal to s to the power for DX here. The derivative of u is equal to one divided by x DX.

And here. If we integrate given because the integral of x to the power for DX, you get v equal to x to the power, five on five, and I'm going to put this into this. So here we have the integral x to the power 4 times, natural log of x DX is equal to u.

That is limb of x then x to the power 5 on 5. Times, natural log of x minus the integral. Then that is x to the power 5, divided by 5. And what is our Du our u is DX divided by x. So this will give us x to the power 5, divided by five times, natural log of x, minus delta.

X to the power, four the x times one out of five and factorize one out of five out x will cancel x to the power, four and x. So x, divided by five times natural log of x, minus 1 divided by 4 sorry, 1, divided by 5 times x to the power 5, divided by 5. Wow. So this will give us x to the power 5 times natural log of x divided by 5, minus x to the power 5, divided by 25. So now we are going to have this as x to the power 5, divided by 5 times, natural log of x squared, minus 2 out of 5. Times x to the power 5. Times natural log of x divided by 5, minus x to the power 5, divided by 5, plus the integral constant c. So this will be equal to x to the power 5 times natural log of x divided by 5, minus 2 x to the power 5 times natural log of x divided by 25 plus 2.

Um, oh, here we have 25 here. Yeah, that is 25 and 5. So we have 2 on 125 x to the power 5, plus 7 as our final answer.

Wow, very nice question with the 9th solution. And I think the next question, find the integer natural log of 2x, plus one, yes for this one, how. Do we use the IETA? This is a composite function. You see that they have natural log of this.

Wow. And then we have an alternate function inside their natural law. So what do we do? We first do the use u substitution.

First to break this one down to break this one down. So let's say that if we take u equal to two x, plus one, we are not following this. You want to uh break this one down four, then we are going to have d u, t, equal to 2 DX. Well. And now d u divided by 2 is equal to the x. So this one will tend. To the integral natural log of u times, d u on two, wow.

And now we can factorize one and two out 1 and 2 times integer, natural log of u. D u. Now for this one, we will use the eyelids because we have u d u. So here, what do we take for our u? So now that we have u here, we are not going to write integral u. D, n, equal to u. V, minus integral then u I'm going to take, uh, my f and g. So here, the first one will be integral f, d, j, equal to f, z, minus delta, then DF so that's, what I'm going to take, so now we will take f. To be equal to natural log of u, and then the gain to equal to, um that is what we are going to do now.

Sorry, the j, equal to d, the g, equal to d. Right? So here we have our f prime to be equal to one divided by u. And if we integrate the gen and integrate the u, I will have j to be equal to u. Wow, very simple as that to integrate this you get that. Now we are good to go. So what do we do? We have the integral f that is natural log of u, the g, as d u to be equal to f. Sum f is natural log of u. Then d is u.

So. I'm going to have u natural log of u, minus k, integral j that is u. And what is the f? The f is one on u. You get it. So it times one on u. Do you as you are putting in here, never as simple as that?

So this will give me u times natural log of u, minus the integral d u and is the same as u times, natural log of u minus u, plus the integral constant c. I think the last question that I saw, I've got the interior constant. Yes, I got it so let's move on check how I did this here. I've used u substitution here. So. That is why I value this formula. So I've used f and g in place of u and v. So don't be confused with it.

Non-Confused briefly with respiration, breathing is the movement of the chest that allows the intake of oxygen and output of carbon nanowire by respiration is different from us. I like that for science people unless my Sendai under mathematics. Okay. So here we took f to equal to natural log of u. And then we took the j to equal to d u.

If we differentiate f with respect to u, we have DF that is f. Prime DF to be equal to d, one, very simple as that. So that is why we substituted it here DF and here, the j equal to if we integrate, we get j equal to g. So that's like that. So let's do our substitution we are going to do that you are going to undo the u substitution here. And remember we have half here already. So we have to use the half. Now you are going to have this to be 2x, plus 1 times, natural log of 2x, plus 1, minus u and u. S, two x, plus one, plus integral, constant c. And we have half multiplying.

It so let's put it here, and then we put the half there. So this one is now going to be half of 2x, plus 1 times natural log of two x, plus one, minus two x, plus one, oh, . Let me put it in brackets constant c. Wow, that's. Our final understanding as our final answer, oh, we can write all of this under a common denominator. 2X, plus 1 times, natural log of 2x, plus 1, minus 2x, plus 1, all divided by 2. Wow, very, very nice. That's.

That is it plus the integral constant c. Thank you for watching and please. Subscribe to my channel bye.