integrate-each-of-the-given-functions-int-frac4-d-xx12x-12-3

Question

Integrate each of the given functions.
$$\int \frac{4 d x}{(x+1)^{2}(x-1)^{2}}$$

Vikash Ranjan · Accepted Answer

problem it is given integration for the X upon x plus one whole Squire, X -1 or a square. Now this is written as for integration Went up on four and two, X plus one Plus one upon 4 and we x plus one whole square -1. A bond forward into X -1 Plus one upon 4 into X -1 whole square the X. That is equal to A landmark, X-plus one -1 upon x plus one minus Ln mode, X minus one -1 upon X -1. Oh, this is equal to Ellen more X plus one upon x minus one- went up on X -1 minus one upon x plus one plus C. So we have this as the answer to the problem.

Amy Jiang · Answer

Okay, so we want to use use up here. So why don&#x27;t we let are you equal are, um, larger polynomial, that is, Let&#x27;s let it equal X squared plus two x. But it&#x27;s what you know, if we take the derivative of that, you get to acts plus two and Indy X, So notice that we have two x puts to here and a DX so that&#x27;s equal to do you. So now let&#x27;s replace what we have. So we have do you over u to the power of four. Okay. So look to your right that as a negative exported so that you to the part of negative four. Do you? And I want to take her in a girl So he gets, um You took out negative three over negative three. Let&#x27;s see. And now, um, what&#x27;s your place? Are you with our X? So we have X squared plus two minus four to the part of negative three over the next three. Your name, plus C

Sid Wan · Answer

Okay. Finds an integral off That&#x27;s over full. Tend Actual full square the ah, that&#x27;s C If we take at our four square you to be that he was Explorer over sixteen D&#x27;You equals axe over Ate the X so we can run this a cz This guy is to see you, Tad. You too. Do you on your hands? You Because ah, well two in front collective national off cause I you we&#x27;LL see if you plugging you to be x square over sixteens that give to natural log because I asked Square over sixteen c, yeah.

Ramzi Deek · Answer

So we got into grow off X over X to the four plus two X squared plus one extra before plus two x squared plus one DX. So, uh, this is the same as saying to girl X over X squared plus one squared DX. So let&#x27;s take you as X squared plus one. Do you over DX as two x So do you Over to X is equal to D X. All right, moving on from there. So you got integral off X over you squared times, do you over to X so X is cancel. Take half outside so half into goal you to the negative too, do you? So that becomes half times you to the minus one divided by minus one. So minus half times, uh, one over you remember you is X squared plus one. So we will get our final answer as minus one over two into X squared plus one plus constant C

Robert Daugherty · Answer

7.1 36 The integral X over X to the fourth plus two X squared plus one. When you see quadratic, you factor them. I don&#x27;t see quadratic, but I do see something that is in quadratic form. So if I look at this, um, if you look at this middle term to half of that is why you square it and you get one. This is a perfect square. This is X squared plus one squared is what the denominator looks like because when you square that you get X to the fourth plus two x squared plus one so I can rewrite this integral as X over X squared plus one squared DX. Now what you&#x27;re always looking for when you&#x27;re trying to integrate it, Do you see a function? And it&#x27;s derivative. Well, if I were to let u equal x squared plus one, that is my function. It&#x27;s derivative. It is going to be two x d x so I can see if I had a two here and a 1/2 in front. Then I see a function and it&#x27;s derivative all side by side. So this turns into 1/2 and then when you look a two x dx, that&#x27;s d U. And then I&#x27;m left with one over. You squared. And so this is just the interval. 1/2 you to the minus two, do you? So the answer here is gonna be 1/2 you integrate this. It would be you to the minus one. Divide by minus one plus a constant of integration. So this ah, answer is minus one over to you, plus a constant of integration. Now, that&#x27;s in respect to the variable you. But you is equal to X squared plus one. So this final answer is minus 1/2 X square plus one plus constant of integration.

Matt Just · Answer

I don&#x27;t know. We have the definite integral zero two four of minus X to the three halves plus x to the one half d x. I could start by finding an anti derivative. So we have so this minus Khun stay are here. And then the entire derivative is so we add one to the exponents We get X to the five hands over five hands plus x to the three halves over three. House evaluated from zero to four. Okay. And so four to the five halves We&#x27;LL take the square root and then take the fifth power. So it&#x27;s two to the fifth, which he be thirty two over five halves plus okay. Squared two cubes of eight over three halves and then minus this evaluated your over that Syria. Okay, so, uh, this is minus can multiply by their sim prickles of both the numerator is get doubles at sixty for over five plus sixteen over three and so we need to get a common denominator. I mean, we could leave it like this, but if we do get the common denominator of fifteen, you should get negative. Six fifty six. Divided by fifteen is the final answer

Problem

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

Problem 14 Easy Difficulty

Integrate each of the given functions.
$$\int \frac{4 d x}{(x+1)^{2}(x-1)^{2}}$$

Answer

$\ln \left|\frac{x+1}{x-1}\right|+\frac{2}{x^{2}-1}-\frac{1}{x-1}+C$

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Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus

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Kayleah T.

Samuel H.

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$\ln \left|\frac{x+1}{x-1}\right|+\frac{2}{x^{2}-1}-\frac{1}{x-1}+C$

Basic Technical Mathematics with Calculus

Grace H.

Kayleah T.

Samuel H.

Joseph L.

Precalculus Review - Intro

Functions on the Real Line - Overview

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

Grace H.

Kayleah T.

Samuel H.

Joseph L.

Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus