integrate-each-of-the-given-functions-int-frac6-p3-d-psqrt9p2-3

Question

Integrate each of the given functions.
$$\int \frac{6 p^{3} d p}{\sqrt{9+p^{2}}}$$

Vikash Ranjan · Accepted Answer

in the problem we have been given integration 19th six P G D P Upon Uvarov nine Plus Pay Square. Now for solving this problem will assume big was 3 10 theater dip equal 36 square theater data. To no This is written as six. Integration 27. Thank you to to into three into six Squad Theatre. The theatre upon Fruit over of nine into 1 plus dan squad tita. Now this is equal to six integration 81 Tan Que Tita In the six sq Tita de Tita upon three 60 to This is equal to 1 62. Integration Don Quixote to sick tita. The tita now further shelving this equal stoop 162 into integration Dennis square theater in Bhutan tita sake theater to the theater. Now late safety deposit key odd 60 to Dante to Did you take was duty Further we have this equals two 1 62 into integration six quarter to -1 into tan theta sick tita lita. This is equal to 162 into integration Do you Square -1 Ditty that is equal to 162 into Due to the Power three upon 3 -D. Now this is equal to 162 into Sick. You teed up on three minus seek tita Further we have this equals two 162 upon three into one plus piece choir upon this is Squire, So peace quite upon nine To the power of 3.2 -162 into one plus He is quite upon 9 to the power of half Further this equals two This is 54 Rather this is 54 should be written like this so it is 54 Into nine plus piece choir upon nine To the power one and 2 nine Plus piece choir to the Upon 9 to the power half this becomes this that minus this is 54 into route over off one plus P esquire plus C. This is equal to twice Into nine plus piece choir in the root over of nine plus P square minus 54. Underwrote one plus PS choir plus C. So we have this as two into 9 plus B sq into whatever off nine Plus piece choir -64 Under Road one Plus piece choir plus see so this is the answer to the problem.

Aishwarya Krishnakumar · Answer

problem were asked to use substitution to find this indefinite integral. So we&#x27;re gonna let you equal this. That&#x27;s inside the parentheses. It could be P puts too. And if we take the derivative of both sides, we get the dura tive derivative of P is just one. So it would be one times deeply now to do the actual U substitution. We would do the integral of you to the six. But then since we have a P left here, we need to do something. So we have this DP We&#x27;re here. We have this over here, but since we have a leftover P, we&#x27;re going to do you minus two equals p. You know, we have a value for P. So you minus two and then we know that d p is equal to do you and we&#x27;re doing this because we want the same letter. Basically, we don&#x27;t want any peas. We only won&#x27;t use. So now you can multiply that out and you would get you to the seventh minus two you to the six. And if you take the integral of this, you would get you to the eighth power over eight minus you to the seventh power were two times one over seven. It was C, and you can&#x27;t have any use in your answer. So you have to plug in P plus two in replacement of you. So keep us too. To the eighth hour over eight, minus two over seven times p post, too. To the seventh hour. We&#x27;ll see. And this would be your final answer.

Matt Just · Answer

Okay, so we have another integral here. Your girl From three to six of dysfunction. X square it X squared plus three. Why d X So why were just thinking is being fixed for announce. Our final answer will be a function of why and now we see we have X squared plus three. Why? And if I differentiate this I&#x27;m going to get our with respect to X I&#x27;m going to get a constant multiple effects. So that&#x27;s our hand to let you be X plus square is to apply. Differentiating with respect to X gives us two x the ex wise constants and just goes away. So x T x is one half to you now I can make her see institutions. So in X is equal to three we have you is nine plus three. Why exit six New is thirty six plus three. Why? And then this is just three U that we have this factor of one half when we substitute in for to you. Okay, so anti derivative of squared of you. You know, the one half is just, uh okay. You to the three house divided by three house and then evaluated from nine plus three. Why two thirty six plus three? Why? And so we have one half dividing by three halves that&#x27;s going to be one third. And then we&#x27;LL evaluate you at the upper limits and then subtract evaluating at the lower limit and that sort of thing.

Vikash Ranjan · Answer

I have a digital before be over nine b, you people. Is there good news to you? Okay. Digging on here a problem? It is terrifying. On lower limited, they become How nine do you over you that people&#x27;s to our 95 Give you the power minus one over to you For though we have this equal school, you know to do You know what you do, you hear? This is nine. But this do this to dance. It&#x27;s therefore we have for our 25 minus a door nine their equities minus tree people&#x27;s too, which is the answer to the

Xiaomeng Zhang · Answer

So by the rule, we know those one dress ico chew Grand knows three p and here is an intro role. When of two and four. So that is my nose. Three times four miners, my nose three times minus two that a consumer miners 12 rhinos, six becomes human.

Aishwarya Krishnakumar · Answer

so for problem 21 were asked to find the definite integral. And to do this be are going to you the substitution. So working about U equals whatever is inside the prank sees. So you equals fine. Why weren&#x27;t in something And we&#x27;re doing that because it&#x27;s a function inside of her function. Now, when you take the derivative of both sides, you get do you equals 54. Why? To the fifth power do you worry and you can go ahead and move the 54 to the other side would be. Do you over 54 equaled? Why? To the fifth hour delay. And as you can see, we have a why to the fifth, do you? Why, in our original equation? So that comes out handy when we do the actual U substitution. However, before we do that, we have to notice that the bounds here are written in terms of why and you have to change that because we&#x27;re doing the U substitution. So we want it in terms of you do that. All you have to do is plug it into this equation. It would be nine times for the upper and ago you would do nine times zero to the sixth, power minus seven. And for the lower integral, you do nine times negative one to the six hour minus seven. And by doing that, you get the upper limit to be negative. Seven. Let me move that and you get the lower limit to be two, and then you can go ahead and do the U substitution. So we have you over here, and that&#x27;s raised to the third power. And then the green circled our do you over 54. So do you. Over 54 and you can go ahead and move the constant to the front. And in this case, the constant is one over 54 and now all you have to do is take the integral. So then I go for you to the third power would be you to the fourth power times one or four. And this would be evaluated from negative 7 to 2. And don&#x27;t forget about the one over 54. So now what this means is you plug in the negative seven into you, and then you subtracted When you plug in to into you. Let&#x27;s go ahead and do that one over four. Time to plug in negative seven and two. You do. Then you subtract the lower bound from so he called in to. And if you plug this into a calculator or if you do it by hand, you get he very big attraction we&#x27;re turning for. And this is the same thing as 11.0 0.4 So this is the answer.

Problem

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

Problem 16 Easy Difficulty

Integrate each of the given functions.
$$\int \frac{6 p^{3} d p}{\sqrt{9+p^{2}}}$$

Answer

$2\left(9+p^{2}\right) \sqrt{9+p^{2}}-54 \sqrt{9+p^{2}}+C$

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Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus

Grace H.

Catherine R.

Caleb E.

Michael J.

Precalculus Review - Intro

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$2\left(9+p^{2}\right) \sqrt{9+p^{2}}-54 \sqrt{9+p^{2}}+C$

Basic Technical Mathematics with Calculus

Grace H.

Catherine R.

Caleb E.

Michael J.

Precalculus Review - Intro

Functions on the Real Line - Overview

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

Grace H.

Catherine R.

Caleb E.

Michael J.

Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus