integrate-each-of-the-given-functions-int_1e-frac3-d-uuleft1ln-u2right-3

Question

Integrate each of the given functions.
$$\int_{1}^{e} \frac{3 d u}{u\left[1+(\ln u)^{2}\right]}$$

Tyler Moulton · Accepted Answer

we want to evaluate the following definite integral. The integral from one to E three D U over U times one plus Ellen&#x27;s where&#x27;d you? This question is testing our knowledge of integration techniques in single variable calculus in particular testing our newfound ability to use inverse. You&#x27;re gonna match up integral. So we have to in such integral are left integral. One is integral to you over a square and a few square equals arts. And you&#x27;re a pussy on the right to states the integral to you over a squared plus B squared is one over A. Are you over april? See we see that are integral match in the form of two. So we have to now figure out you do um A are you as L M U R D u S t U over U and R equals one. So are integral has an extra factor of three that we must carry over to the solution. Thus we evaluate the integral as three times in trouble. Do you ever you Over one. Radio is three arc tangent. Ellen you from each one or Three times. They are changing 1 -3 times, are Changin zero, so in parentheses as follows.

Aishwarya Krishnakumar · Answer

So for problem 19 were asked to solve his definite integral and to do this work. Gonna just move this to the numerator. That makes things easier for us. So we&#x27;re gonna go ahead and rewrite this inter goal to eat before you minus and remove this to the top. I love you to the negative to power. Now we can just go ahead and take the integral. Just look at a part by part. So the first part is eat before you to the integral of that is just eat before you times one over four and then look at the second part, which is you, plus one to the negative second power. And that would be you, plus one to the negative. First power over negative one who&#x27;s with minus sign. And you have our way, this whole thing from 1 to 2, because those are the bounds. So now you just have to plug in to into you. So we&#x27;re going to do that. I hear. So 1/4 plug in to into you. So, hee, four times too minus you. Plug in to again. Two plus one to the negative press power over minus one and then you plug in one, you subtracted. That would be 1/4. So now all you have to do is maybe do a little simplification. So to both one that&#x27;s three. So it would be 1/4 p to the A, and then the minus minus cancel out. So it would be plus won over three. So I&#x27;m just moving the whatever&#x27;s to the power of negative one to the bottom and minus 1/4 e to the fourth. And then this would be a plus. However, there&#x27;s a minus out here. It would still be minus. And then you move one path one to the negative one. Power to the denominator would be won over one plus one, which is to So should we simplify this even more you get 14 You did that, eh? Minus 1/4 either the four minus 1/6. And if you plug this into a calculator good. 731 point for two

Xiaomeng Zhang · Answer

So that&#x27;s why you have todo into grow off me to the power off for you Minor. So injured grow Global goddess, Do you hear off while over u plus one Squire. So for 1st 1 that it goes to eat the power off for you over four while too. And for those? Well, we used a substitution. We&#x27;ll see here, you make it pass new class one and those over anything change our were in here. Actually, now that you&#x27;re Avital often you plus one dressy co +21 So we don&#x27;t need to do any other change here, so I can&#x27;t say it. That&#x27;s why you go to speak the power off eight miners each of the power for over for and with us. When he comes to run over, he be he That&#x27;s the one We know that a cow too long and absolute value off t. So yeah, How me to the power off eight minor C to the power of four or four minus law on street last law and true reach approximately. You call to several hundreds or on 31 pawn for two streets

Xiaomeng Zhang · Answer

so because we know that the roto off long abstract Seiko So one over accessories a substitution here. So we the b relative, are long access because to this one overact So here, because long over acts and also in nature changed us to arrange Yes, the long on dressy coats of zero there are become you meet you and bad address ICO to you, Squire over to two euro Here is long choose So we&#x27;re how Ron chew to the power off to over two minus zero. And that one ICO too. The rope on to 40 to truth.

Amy Jiang · Answer

Okay. So, looking at their own six point on, we have the inner girl of I provide close begins, not equal to the natural laws of that value of hyperbolic tangent of X over two. And this is it all the way to add our endpoint. Okay, so that gives us the natural laws of tangents. Uh, the natural laws of degree over to I didn&#x27;t minus the natural log of absolutely attentions. Uh, the natural log of to over to Okay, so this is our solution.

Michael Jacobsen · Answer

for this problem, we&#x27;re going to be finding the anti derivative of this expression that we see displayed here. This is a quite a complicated combination of algebraic expressions. So what we&#x27;re going to do is start with a U substitution so that we can reorganize this integral into something that&#x27;s easier to work with. Let&#x27;s let you be equal to the ordinary natural log of X. Then the differential of you would be one over X DX. This is going to allow us to rewrite the integral as natural log X squared is now you squared. We have exploited the denominator, but we have to separate ah one over X to combine with DX in order to produce our d&#x27;you. We do still have one problem. We have to do something about this variable X since everything must be written in terms of you. So what we&#x27;ll do is go to this original substitution and express it as follows Well, right. You to the U is equal to you the power of natural Log X And we have a cancellation here then so that you to the U is just the same as X. Now we can go back to our integral and it becomes anti derivative of you squared divide by X, which we found to be each of the U times one over x dx But we found that to be equal to do you If we rewrite this integral into something more friendly to work with, it&#x27;s the integral of you squared times e power negative you do you? So this is that reorganized injure grant that we can work with The method of integration by parts works very well in this situation. So let&#x27;s let a very bold w denote you squared. Then it&#x27;s differential is to you times do you. Next we can let DVB equal to either the power of negative you. Do you? This would imply that V is negative either the power of negative you where this extra negative sign is per the chain role then going back to our integral. This now turns into W times V, which is negative. You squared either the power of negative you minus the integral of v times d. W. We have a constant of negative one times two so we can rewrite this as plus two and right in you times either the power of negative you Do you? Then at this stage, are indefinite. Integral can be once again evaluated with the exact same method of integration by parts. Let&#x27;s let w b you This time the differential is dw equals do you and likewise Devi will once again be equal to either the power of negative you d u So v is once again negative to the power of negative. You going back to our equation down here will then pick up negative. You squared either the power of negative you plus two times that group found by taking w times V, which is negative you times e to the power of negative you. Then we subtract away the integral formed by taking v times d w. There&#x27;s an extra negative one factor here so we can convert this to a positive if we like and obtain e to the power Negative. You Do you notice that we&#x27;ve already done this evaluation twice the anti drift of of either the power negative you Is this quantity here negative of either the power negative you. So this tells us that Oliver evaluations are complete on this next line we have negative. You squared U to the power of negative view minus two. You have the power negative You through a distribution here to here. Another distribution here to here will give us two times this result. So that would be a minus two U to the power of negative you plus a constant integration. See, we could do a couple things at this stage if we like. We can factor out negative to the power negative. You leaving behind you squared plus to you plus two plus a constant C. The next step we must take is to return to the original variable X But that could be done by using this Originally, you substitution. We have that for example, U to the power of negative view is the same as each of the power of negative of the natural log X. There&#x27;s no cancellation here at this stage. We must first use the property of logarithms that says this coefficient of negative one can become a exponents on X that gives us e to the power of the natural log of X power. Negative one. And just as we saw in the original set up here, there&#x27;s a cancellation with these two functions. Since they&#x27;re a pair of in verses, we&#x27;re left with expire negative one. Or if you prefer one over X, let&#x27;s now state our solution. We have negative of a power negative view, which is one over X times you squared. But that&#x27;s the same as the natural log of X squared, plus to you or to natural log of X plus two plus a constant integration, see, and this is our anti derivative.

Problem

Integrate each of the given functions. $$\int_{0}…

Problem 15 Easy Difficulty

Integrate each of the given functions.
$$\int_{1}^{e} \frac{3 d u}{u\left[1+(\ln u)^{2}\right]}$$

Answer

$\frac{3 \pi}{4}$

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Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus

Anna Marie V.

Caleb E.

Kristen K.

Joseph L.

Precalculus Review - Intro

Functions on the Real Line - Overview

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$\frac{3 \pi}{4}$

Basic Technical Mathematics with Calculus

Anna Marie V.

Caleb E.

Kristen K.

Joseph L.

Precalculus Review - Intro

Functions on the Real Line - Overview

Allyn J. Washington, Richard S. EvansBasic Technical Mathematics with Calculus

Anna Marie V.

Caleb E.

Kristen K.

Joseph L.

Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus