integrate-each-function-by-using-the-table-in-appendix-d-int-frac3-d-x9-16-x2-2

Question

Integrate each function by using the table in Appendix $D$.
$$\int \frac{3 d x}{9-16 x^{2}}$$

Vipender Yadav · Accepted Answer

So here in this problem we have to evaluate an indefinite integral which is integration three dx over nine minus 16 X square. And to evaluate this interval we have to use the formless given in the appendix day off the book. Now let me forced rewrite this integration hours three D x. Now I&#x27;m taking negative sign outside. So this would become 16 X square minus nine. And this and nine can be ridden us three square and 16 X square can be written us four X whole square like this. Mm Now going to use the formula that is given in the appendix city of the book. So I&#x27;m going to use the formula which is integration do you over you square minus a square And this is equal to one over to a natural look U minus a divide by U plus a blessed see where this C is a constant of integration. That will come because it&#x27;s an indefinite integral. And here also this indefinite integral. So plus C would come. Now I&#x27;m taking this three. Outside the integration it would become negative three here. Now we can exactly use this formula to evaluate this integral. So it will be equal through this minus three will remain as it is. Now if you compare this N trickle with the given formula, we can say that here you is equal to four X. So U. Is equal to four X. And do you will be equal to four weeks. And her A. Is a girls too. This is three here. Now if you see this expression we don&#x27;t have four years here. So we want to multiply Edward four and dividing it with four. Now we can use this formula. So this would become minus three. Where Forest this we are taking outside the bracket now 1/2 8 A. We have a 3 to 2 times 36 natural log this is U minus six. So you&#x27;ll be here for X. So for x minus eight we have three divide by A. U plus A. So you we have four X. So this is four X plus a. Three plus. See we&#x27;ll see the concept of integration that will come because it&#x27;s an indefinite interests that I have explained earlier not to further simplify it and finally we&#x27;ll get the answer to be 1/8. This negative finance, it is natural log four X minus three, divide by four X plus three. Let&#x27;s see. So this is a required answer for the given problem. I hope you have understood the problem. Thank you.

Robert Daugherty · Answer

problem. 17. Um, So in this particular section, I&#x27;m looking at indefinitely rose and using integral look up tables. So one of the keys has find a table that you know either in your book or an Internet resource that you like and you&#x27;re comfortable. But it&#x27;s just familiarizing yourself with a table. So when I look at this, I know that I&#x27;ve got to go to the section that organizes things that look like a X plus B. That&#x27;s where I&#x27;ll find the solution to this one if it&#x27;s listed. So in my particular case, when I look at my integral table, um, that&#x27;s what I see here. Excuse me. Not x plus B, but a X squared. This was a square term that was here. Let me back at Rose quick and see if I did that correctly. So yes, So it&#x27;s nine X, and that was a typo on my part. Um, this was nine x squared. So in my particular case, it&#x27;s a X squared. Plus, C is the way that my particular integral table is is organized. So if I look at what I have, then in this particular case, uhm, I&#x27;m going tohave, um X squared plus C, which is what you see there. So X squared plus C to the three halves. And then I see the value for this anti derivative. So according to my look up table, then what I&#x27;ve got is Thean, a girl of one over a X squared plus C to the three halves. DX is equal to X over, see and then square root of X squared, plus C plus. Not to be confused. Ah, constant of integration. So in this particular case, I&#x27;ve got a is equipped in nine and then little C is equal to 16. So plugging it into the formula I&#x27;m going to have X over, um 16 and then the square root of nine X Square plus 16 plus a constant of integration. So that&#x27;s my final answer. X over 16 Radical nine x scripless 16 plus a constant of integration

Matt Just · Answer

Okay, so you don&#x27;t evaluate this integral of a function of two variables. So you go from one to two of next. Why? Cute minus X do you want? So we&#x27;re evaluating this, Integral with respect. Why on we&#x27;re treating exes if expert just a constant. Okay, so the final answer will actually be a function of X. So we need to take an anti derivative of this grand with respect of why so have X. Why did the fourth of before thanks for the stream like a constant minus x y. And then we evaluate this from one to two. So you&#x27;re finally answerable. Plug in two for why this is Y equals two and Michael Spine. So that&#x27;s going to be eight over. Let&#x27;s see to the fourth at sixteen. So for X minus two x, okay. And then that&#x27;s evaluated it, too. And then minus We&#x27;LL have X over for minus It&#x27;s X. Okay. And so this is two x minus. This is going to be negative. Three fourth ex said to X is like eight Exeter for plus three Exeter, for that&#x27;s gonna be eleven words. Next

Patrick Adelstein · Answer

Okay, We&#x27;re gonna do the indefinite integral. Uh, this function of X one over quantity 16 plus nine squared the three over to power. And so this looks very similar to one of the rules in our table Vina goals. We have to do a little substitution in order to make it look exactly like one of the rules. The first thing on a show is exactly why it reminds me of a rule in the table of inter girls. Because in the bottom we have a content being squared plus something with X being squared as well. All to the three house power and the substitution we need to make is that function of X being squared that three x So, as always, after I choose my substitution, we differentiate it. And in the inner Groll, there&#x27;s no three d X. There&#x27;s just a DX, so I&#x27;m gonna move the three over, divide both sides by three so that I can replace the D X in integral with 1/3 dear. All right, let&#x27;s do that. So it&#x27;s in a Groll. We&#x27;re still gonna have quantity for squared Plus and I&#x27;ll replace that the reacts with you now still to the three halfs power, and I&#x27;m going to replace the D X with the equivalent 1/3 d&#x27;you. We can pull Constance out of the integral. Great. Now this looks exactly like a rule we have in our table of intervals one over a squared plus variable squared to the three house power. So use no the color to represent that I am simply applying the appropriate role from the table of angles. It&#x27;s gonna be 1/3 times. That was 1/3 time&#x27;s the entire integral. Okay, we get variable plus pretty common thing. Natural log of our variable plus square root of the A squared plus the variable squared closer natural log. Close that big bracket and don&#x27;t forget constant integration. All right, And then next we&#x27;re distribute that 1/3 and I&#x27;m going to switch all the EU&#x27;s back. Teoh their ex forms. Well, remember Yukos directs that waas the substitution we made. Okay, 1/3 you, which is the reacts plus 1/3 natural log of three X. We&#x27;ll go ahead and write that foursquare to the 16. Now let&#x27;s see you three x squared. That&#x27;s nine x squared. And don&#x27;t forget the constant of integration

Andrija Isakov · Answer

in this question will be finding the integral of X times two X quick two to the power of X quick. So first notice that we have an X squared was derivative is two X and we have X dx in the expression as well. So we noticed that we have a function inside of function and the inter function also has a derivative present. So that gives us the hint to use substitution. And we looks may substitute you for X squared, which means the value of the integral or the interval can be rewritten as to to the exponents you times 1/2 the year. This, of course, equals 1/2 of the integral of to to new to to the U, which by our integration table by formula equals to the exponents you over lawn of to plus the constant of integration. And so no, we have found the indefinite integral. We found, um, the family of anti derivatives. I would correspond to our original integral. Now we just have to substitute in terms of the variable. We had a beginning X since U equals x squared. You can write this us two to the power of X squared minus one all over long enough to close the constant of integration

R M · Answer

problem. Number 40 Configuration off X A configuration of excess square plus six x over X squared for three. Power to be X, which is equal to integration off X squared plus three over X Squared the block three forward to the X plus integration off six x over X squared the for three for two, the X minus invigoration off three over X squared, both 3/2 and the ex Ah, this can be simplified into integration off one over X squared per three The x both three invigoration off two x over X squared for 3/2. The X minus three invigoration off one over X squared of both three. It&#x27;s great e x, and this is equal to so the first integration here can be sold with using poor Jamuna 32 from Beach T two. Um, so the answer for that is equal, and the second integration can be evaluated using a substitution as you is equal to X, it&#x27;s squared plus one. So you is equal to ah to Rex the X ah, so the integration can be written as integration or one one over you squared you and the third integration can be evaluated using Formula 33 on Page de to. So the answer for the integration for the three integrations is equal to one over the square. Root for three, then in verse over X. Over is growing with three miners, three over X squared, the for three minus X would two X square for three minus 1/2 square root for three, then in verse, X over square root or three process.

Problem

Integrate each function by using the table in App…

Problem 24 Easy Difficulty

Integrate each function by using the table in Appendix $D$.
$$\int \frac{3 d x}{9-16 x^{2}}$$

Answer

Related Courses

Basic Technical Mathematics with Calculus

Related Topics

Discussion

Top Calculus 1 / AB Educators

Grace H.

Anna Marie V.

Samuel H.

Michael J.

Calculus 1 / AB Courses

Precalculus Review - Intro

Functions on the Real Line - Overview

Recommended Videos

Watch More Solved Questions in Chapter 28

Video Transcript

Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus

Grace H.

Anna Marie V.

Samuel H.

Michael J.

Precalculus Review - Intro

Functions on the Real Line - Overview

What do you need help with?

Textbooks

Ask a question

Courses

AI Tutor

Basic Technical Mathematics with Calculus

Grace H.

Anna Marie V.

Samuel H.

Michael J.

Precalculus Review - Intro

Functions on the Real Line - Overview

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

Grace H.

Anna Marie V.

Samuel H.

Michael J.

Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus