integrate-each-of-the-given-functions-int-fracd-rr-ln-r

Question

Integrate each of the given functions.
$$\int \frac{d r}{r \ln r}$$

Bobby Barnes · Accepted Answer

So when we&#x27;re going to find this indefinite ineffable using the substitution, the first thing you want to do is kind of think about Well, what&#x27;s the thing that&#x27;s really making this difficult to integrate? So if I want to integrate our times e to the negative r squared with respect are you might kind of see how this e to the negative are square is giving us issues So you might just first make the guests that you is equal to he to the negative are square And then we can go ahead and take the derivative of this with respect to our to find our gave you We&#x27;LL get you over e r is equal to so taking the door But it will be to the negative are squared will be changeable because remember, we have parentheses right here So this is actually the composition with each X So we&#x27;ll end up with first just e to the negative r squared since the derivative of X urgency to necks So that doesn&#x27;t change And then we have to multiply by the derivative of the inside functions ddr of negative are square and this here will be eats his negative are squared times negative to R. This is You are So you might notice that this looks very similar to what we already have. But the only difference is we&#x27;re being multiplied by this negative too. So what I&#x27;m going to do is I&#x27;m going to divide over by negative too. Bye bye, dear. So doing that gives us d&#x27;You over Negative too is equal to on i&#x27;LL move the r out front So ar e to the r spirit d r And now this here is exactly what we have here. So let&#x27;s go ahead and re vipers has the integral Oh, so I&#x27;ll plug in d&#x27;You over Negative too. You over mobile too. And now we can cool the negative one half out negative one have integral of one do you? And remember when we integrate a constant will just end up with the variable Negative one half you lost See? And now we can go ahead and take our view It&#x27;s here plug it in on it Do so in negative from half heed the negative are square What? And this year will be our indefinitely

Amy Jiang · Answer

Okay, so it&#x27;s starting or thing that in the exponential function or power is negative. R squared. So if we let that be cool to you, then our deal that&#x27;s people to negative. Two R g r And there was here. That&#x27;s RG art is part of our integral. So that&#x27;s solved for RTR. So we&#x27;ll divide by negative to on both sides. So we have Do you over Negative two is able to car are okay, So now that we have our g r and are you what&#x27;s your place or ours before use? Do you get each and power of you and then do you over and they go to Now let&#x27;s take out or negative too. And no, let&#x27;s take her in a row. So that&#x27;s this one house or negative 1/2. Um, you need to do you not see and now lets her place are you so we get each of power of negative R squared plus C

Joshua Silva · Answer

Okay, So the inner girl that we have here to evaluate is the integral of the quantity R squared plus our plus one all multiplied by E T v r D r. And this is a standard in the integration by parts problem. So just to remind ourselves of the formula for integration by parts we have here, the integration by parts tells us that the integral of anything of the form you Devi on Girl of this form can be re written equivalently as you times v minus the integral of VD you and to just get a quick understanding quick glimpse of sort of the visual intuition behind why integration by parts. It works the way that it does. We have here a little graphic which can give you that If you examine it, you can pause a video, take a look. You can see how all the pieces fit together to give us this formula. But moving on to our integral we I want to find a part of this expression to be Are you in another part of this expression to be our Devi And naturally we want to pick a you that will differentiate down to something simpler so we wouldn&#x27;t want to pick the heat of the R because that would just give us another exponential function When we don&#x27;t differentiate it, that would be just the same thing. So in instead of that, we&#x27;re going t take this polynomial here as being Are you? So we write, let you equal r squared plus our plus one then d you will be the derivative of this thing. Times d are so that&#x27;s two are plus one d r Devi, in this case is going to equal you to the e r Do you are? And then RV will eagle by integration Integral of the r d r is just e to the r so And they worked that out Let&#x27;s get on with our integral We have the integral of our square plus our plus one you to the e r g r equals you times v So you is our square plus our plus one and ah v is really are minus the integral of VD You ve again is either the r and D u is to our plus one d are So when we put these together are integral will look like two are plus one times e v e R D r. Now, in order to evaluate the second, Integral will have to perform another application of integration by parts to simplify it a little further for our u will want to pick two Arpels one and then we&#x27;ll have either. The r D r r D v So working that out, we say, let you equal to our plus one. D&#x27;You in this case equals two times D. R and R d v equals e T R D r and R. V Again is the interval of eating the R D r equals each of the R already then. So writing this out we have you Tim&#x27;s free is to our plus one times easier minus the integral of Edie you where v is if they are and you is to d r so putting that all together we have the role of to you we are d r. And then ah, putting this all together again. We can distribute this negative over the sun and get positive the integral of to eat of the r d r. Now this integral is just to eat of the are so we can write you out like that. And then finally, we can add a constant now to simplify our answer. We can collect like terms since this heat of the R appears and all of these factors in all of these terms. And so we just add all these polling no meals together in front of it and get our squared. Plus are minus two are gives us minus r, and then plus one minus one plus two will give us plus two. So that times either they are plus C and that is our solution.

Vikash Ranjan · Answer

he is. No. Yes, Yes. One into a bar. This is even honest. And do you love you my ass deep? I DEA honest You are dressed. Yes, it is even one in my past. So this is it one dude. But my price off and now way have to solve for you. So this is because disease Dia dia, this is Is the guns honest? Mmm. Ms. Yes. Minus last. You This gas is out. So this becomes honest about. See, this is equal to about thesis the fines given from

Kyler Gray · Answer

for problem 14 were given a derivative that we&#x27;re gonna have to use integration by parts and the tabular method to solve. So we&#x27;re gonna make it of the Are our a d v e to the R. Since that&#x27;s extremely easy to integrate, as it will just be you&#x27;re a tr tr. And so on on, then we&#x27;re gonna make our squared Plus are plus one. Are you so well, differentiate that. So we&#x27;ll start with R squared. Plus our plus one on derivative of that will just be to r plus one derivative that will just be two, and then again would be zero. So now we can go ahead and connect all these up Connect, connect, connect and alternates a plus minus plus And there we go. So this is the tabular method. It just saves half into do multiple layers of integration by parts. When we have repeating pieces like either the r and something that&#x27;s just gonna come down to zero. So now we can assemble all of these. So we have a positive r squared plus are plus one. If positive that times either they are, then we have a negative to r plus one times a day they are, and then we have a positive to e to the r. And then all of that will be plus C. Um, And so there we go. That is our final solution.

Problem

Integrate each of the given functions. $$\int \fr…

Problem 21 Easy Difficulty

Integrate each of the given functions.
$$\int \frac{d r}{r \ln r}$$

Answer

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

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Basic Technical Mathematics with Calculus