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JH
Numerade Educator

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Problem 28 Hard Difficulty

Evaluate the integral.

$ \displaystyle \int \frac{x^3 + 6x - 2}{x^4 + 6x^2}\ dx $

Answer

$$
\ln x+\frac{1}{3 x}+\frac{\tan ^{-1}\left(\frac{x}{\sqrt{6}}\right)}{3 \sqrt{6}}+C
$$

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Video Transcript

Let's evaluate the integral first thing we should do is to see if this denominator will factor. We see that in the bottom we can pull out her X Square and you left over with X squared plus six. So for the first term in the denominator, we see that this is a repeated linear factor. X times X Now for this other quadratic X squared plus six. I would like to see if that factors so to do so we checked indiscriminate, Which is he squared minus four a. C. Here B is the coefficient in front of the X. There's no X here, so that zero minus four times of one in front of the Explorer and then Time C, which is six. And that is definitely less than zero. This tells us that the denominator X squared plus six term will not factor. So for the first term X squared, we can use what the book calls case to repeated linear factor. And then, for the last term, Explorer Plus Six will use with the book calls Case three CX plus de over X squared Positive. Now let's take this equation up here and multiply both sides by that denominator on the left. When we do so, we should get X cubed. Six X minus two No. On the right hand side, we have a X X Square plus six B that's times X squared plus six and then CX Clos de and that gets multiplied by X squared. Let's just go ahead and expand this right hand side. X cube Si se x be ex players six me see execute d X squared and then now has combined like terms. We see that we can pull out of X cubed a pussy x squared, rusty plus X and x just six a m. X of the x And then the constant term is just at six b So we'll have four equations here we see that a plus e must equal one because those are the coefficients in front of the execute terms. On the left we see that there's no exterior, no X square term. That means that b plus he must be zero in front of the X On the left we see a six. It's on the right. That must mean six a equal six and finally the constant term on the left is minus two on the right in six b. Those must be equal. Let's write and solve this for my four system on the next page, our equations a policy is one B plus zero six a equal six six B minus two solved after B solve this equation for a and then we can go ahead and plug in R and d values to solve for C N. D. C is one minus. A is one, so we just could zero proceed. B equals negative d or let me write. It is day equals negative. D equals one third. Now let's plug in these values for a, B, C and D into our partial fraction to composition. And we have one over x minus. Wondered. It's clear that was for B and then see was zero and then d was also He was positive one third. Now we could go ahead and write. This is three separate minerals. You see the second integral there. You can use the power will for that one. And in the last integral pull up the wonder. And technically, I should be putting the DX is here dx d x No. For this term we know that's just natural. Log Absolute value X Here. This will be negative one third times negative one of Rex after using the powerful. So that's just one over three X And then for this last integral. You might remember this from a table. If not, we'LL do it turns up, let's come over here For that X equals Route six Cantina. Then D X equals Route six. C Can square data either. So we have wondered in a rule D x and then on the bottom X squared. So that's six hands where data and then plus six. I'll be fat around that six and then we know that tan square. Plus he can't tan squared. Plus one is he can't swear. So these terms are equal. We could go ahead and cross those off. We get one six over eighteen. That's the three times in six in the bottom in a rule data. So that's one six over eighteen of data, and then we can solve for theta by taking our tricks up over here and just solving that for theater. So we have Tien Data equals X over the radical six. So Dana equals tan in verse X over radical six. So let's write that over here. Radical six over eighteen and then Arc Tan X over radical six. And that's the third interval. So the last thing to do is write to some of these threes terms, and so let's go to the next place to do that. So we had the first integral natural log. Absolute value. Second Rule, one of the three X and then for the last integral was required to trips of one six over eighteen. And then we had heartening X over radical six and then don't forget their constancy of integration, and that's our final answer.