integrate-each-of-the-given-functions-int-frac4-x221-x6x2x-3x4-d-x-3

Question

Integrate each of the given functions.
$$\int \frac{4 x^{2}+21 x+6}{(x+2)(x-3)(x+4)} d x$$

Vikash Ranjan · Accepted Answer

in the problem we have to find the the girl which is integration for its Squire Plus 21 x plus six dan X-plus two into X -3 and two. X plus four dx. This is written as Forex Squire plus 21 x plus six bon X plus two into x minus three. And the X-plus four that equals two Upon X-plus two plus B upon x minus three plus C upon X plus four. Now this is written as four X Squire plus 21 X plus six that equals two in two x minus three into X plus four Plus being two x plus two In the X last four Plus seeing two x plus two In the X -3. Now by using his gentle method we obtain the value of A. That equals two, two be equal to 3 & C Equal to -1. Therefore integration becomes two upon x plus two Plus 3. -3 -1 upon x plus four Deluxe there four days twice. Ln more X plus two Plus three. Landmark X -3 minus Ln more X plus four plus C. So this is the final answer.

Tom Greenwood · Answer

Okay, we have this in the Kroll, and the first thing we have to do is long division because we have the same power, the same degree in the numerator as we do in the denominator. So we have to take X squared minus three X plus two, and we have to divide that into three x squared plus four x minus six. And so that&#x27;s going to divide three times because three X squared, divided by X squared, is three. So now multiplying the divisor by three gives us three X squared minus nine X plus six, and then we subtract those and we end up with 13 X minus 12. So this integral actually becomes the end girl of three plus 13 X minus 12 over our X squared minus three X plus two d x. So now we have to do our partial fraction decomposition the 13 x minus 12 over the X squared minus three x plus two. And hopefully you noticed that the denominator factors and one of the factors is X minus two. Another factor is X minus one. So we have a over X minus to plus B over X minus one as the form of our decomposition, and we multiply everything by the common denominator to clear out the fractions. So 13 x minus 12 equals when we multiply the fractions thes denominators, air going to reduce So we&#x27;ll just have the numerator times the other denominator. So that will give us a X minus a plus B x minus to B, and now we can equate coefficients. We&#x27;ve got 13 x 13 equals x plus B x +03 13 equals a plus B and then equate are constants. The negative 12 is gonna equal the negative a minus to b And now pick whichever method you want to use to solve this system of equations and we end up with a value of a equals 14 and b equals negative one. So that lets us rewrite are integral one more time as the integral of three plus 14 over X minus two minus one over X minus one DX. And now because we just have addition and subtraction weaken. Just integrate each of these individually, Uh, and a girl of three is just three x integrating the fractions. We can pull the constant out in front if we like, make 14 and be one t X over the X minus two. So now these air in the form of the one over you, d you so that will give us 14 times the natural log of the absolute value of X minus two minus the natural of the absolute value of X minus one. And then we have our pulse, See, because it&#x27;s an indefinite integral there. And a final thing we can do is use our rules for algorithms to combine those all together. So we end up with three x plus the natural of the absolute value. 14 becomes an exponents on the X minus two. And the furnace attraction means we&#x27;re dividing by the X minus one all inside the absolute values plus C. And there is our integral. And if I don&#x27;t touch the touch pad around writing, that makes it

Tom Greenwood · Answer

for this problem. We first noticed that the degree of the numerator is bigger than the degree of the denominator. So that means we have to do long division in order to simplify this and get it down to a form that we can actually work with. If you&#x27;re doing the long division by hand, you have the Z squared plus Z minus six outside your box and the two Z to the third plus the squared minus six Z plus seven inside. Now, I&#x27;m not gonna go through the long division. There actually are websites that you can go to that will do. Is it division for you? So what we end up with here is when we divide this, we get to Z minus one a czar quotient, and we end up with a remainder from our final subtraction of seven Z plus one. So this integral becomes the integral of to Z minus one waas 70 plus one over the Z squared Pause Z minus six. Oh, do you see? So now we have to take our fraction and we have to used the partial fraction decomposition for that So that seven Z plus one over the Z squared plus Z minus six. The denominator will factor and one of the factors is equals three and the other factor is Z minus two. So we end up with something over Z plus three, plus some other value over C minus two. Those added together to give us this fraction and now we multiply by the common denominator to give it of the fractions and figure out what those two numerator are going to be. So we have seven Z plus one on the left hand side and on the right hand side, each denominator will reduce, leaving us with the numerator times the opposite denominator. So eight times the Z minus two will be a Z minus two a b times. The Z Plus three will give us BZ plus three B and now we can equate the coefficients of our terms. So, equating the Z terms, we end up with seven equals a plus B and equating the constant terms we have one equals negative to a plus three b. And whenever you solve that system of equations, they end up with a equals four and B equals three. So now this becomes the angle of two z minus one Waas four over. Z plus three plus three over. Z minus two. Easy. So now we can just integrate each of these individually and a girl of A to Z is going to be Z squared in a girl of minus one will be minus Z and integrating our fractions. We could pull the constant out in front and we would have this in the form one over you T U. Because derivative of each of the denominators is just gonna be easy. So that gives us four times the natural log of the absolute value of Z plus three plus three times the natural log of the absolute value of Z minus two. And we&#x27;ll have our plus arbitrary constants c there. And then we can combine our algorithm expressions together using the rules from algorithms. So we have Z squared minus. Is he plus the natural log of the absolute value uh, the Z post three to the fourth Power times. The Z minus two to the third power, plus C

Aishwarya Krishnakumar · Answer

for problem 19 were out to u substitution to find this indefinite integral. So we&#x27;re gonna let you equal the denominator. U equals X to the four plus for X squared plus seven. Now we take the derivative of both sides, maybe four x cubed eight. Thanks D x, and you can simplify this by taking out a two X. Actually, you could take it a four x my bad B X squared close to detox and then you can divide both sides by four. It would be do you over four equals x times X squared clothes two times d x So now as you can see this, if you foil it, this would be right a year x cubed plus two x, which is the same thing as a the numerator of the original equation. So that makes it easy when we rewrite this interval. So it beat the integral of won over you because of this. And instead of the three X, we would have do you over four and you can move the constant the front would be won over four, which was the constant and then the integral of you to the negative first power and in a girl of one of her you is out and of the absolute value off you. And don&#x27;t forget the plus C, and you cannot have a you in the answer. So we&#x27;re gonna plug in this You So, Helen, uh, next to the floor waas four x squared plus seven post C and this is the answer.

Amy Jiang · Answer

Okay, so let&#x27;s start another thing that we have a fixed in power for here, and then we&#x27;ll go toe that we get extra party, which is what we have in our numerator. So what do we let you equal our denominator? Okay, I know. Let&#x27;s take our derivative or hit, and then we get the nuts. Okay, so we have an X cubed plus two X. So we confront their out of four here. And that leaves us exactly what with what we once so would you divide by four on both sites? Okay, I know it&#x27;s here. That&#x27;s gonna be RG you or do you over four in this portion here, that&#x27;s are you. So we get the integral of do you over four and in you. So it&#x27;s factor out a woman. Okay, so we get 1/4 in a little of do you over you and are integral. That&#x27;s you. Cool, too. That&#x27;s critical to Ellen of you. I see. To be an Ellen of you, that&#x27;s extra part four parts for X squared plus seven and in policy

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.

Problem

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

Problem 16 Easy Difficulty

Integrate each of the given functions.
$$\int \frac{4 x^{2}+21 x+6}{(x+2)(x-3)(x+4)} d x$$

Answer

$2 \ln |x+2|+3 \ln |x-3|-3 \ln |x+4|+C$

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Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus

Catherine R.

Anna Marie V.

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Precalculus Review - Intro

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$2 \ln |x+2|+3 \ln |x-3|-3 \ln |x+4|+C$

Basic Technical Mathematics with Calculus

Catherine R.

Anna Marie V.

Heather Z.

Samuel H.

Precalculus Review - Intro

Functions on the Real Line - Overview

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

Catherine R.

Anna Marie V.

Heather Z.

Samuel H.

Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus