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Numerade Educator



Problem 33 Hard Difficulty

Evaluate the integral.

$ \displaystyle \int_0^1 \frac{x^3 + 2x}{x^4 + 4x^2 + 3}\ dx $


$$\frac{1}{4} \ln \frac{8}{3}$$


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Video Transcript

Let's evaluate the given Integral. First, we should see if we can do a partial fraction The composition. Now, looking at this denominator, this is really a quadratic in disguise. If you want, you could think of X to the force as ex players where then we could factor. This is X squared plus three x squared plus one. So we have to quadratic ce and now we would should check whether or not these quadratic faster. So let's look at their discriminative b squared minus four a. C. So for the first polynomial, the first quadratic over here circled. We see that is one D zero and we have three equal. See, So for this one will have zero minus four times one times three. That's a negative number. So has pointed out in the section in the reading. This quadratic will not a factor. Similarly, over here we have B A zero minus four and then a is one and see is one what? So we have negative for less than zero. So that means that this quadratic also doesn't factor. And now, using what the author were called case three non repeated quadratic factors and then for the second one. We have CX plus D and then X squared plus one. Now let's take both sides of this equation up here and multiply it by this expression, this denominator in the left side after doing so on the left. But on the right, we have X plus B Times Explorer plus one plus CX plus Dean, and then explore blustery. Let's go ahead and expand out this right hand side as much as we can over a X cube B X Square X B and that for the second factor, we have C X cubed D X flair and then we have three C X and then three d. Let's go ahead and factor out like terms. So here, let's take on X Cube factors that Oh, and then we left over with a policy. Then let's do the same for X squared Rusty politics April through see and then left over with Be just ready. Now let's look at the coefficients on the left and on the right under left. We see that there's a one there in front of the X Cube. That means on the right a plus. He must be one we see that there's no X squared term on the left. That must mean that the X Square term on the right, the cowfish it must be zero. So B plus three equals zero. The coefficient on the left in front. The ex is the two on the right. It's a plus three city, so those must be evil. And then under left, we see that there's no constant term without X value. It's on the right. This constant term with no X must be zero. So it's right out and solve this four by four system of equations for A, B, C and D. And when we do, we could replace the Inter Grand with this expression up here in the red on the right, and then we'LL integrate. So we have our system. No, it causes one. We also had a plus three c equals two b plus the equal zero and then B plus three d equals zero. So, for example, if I just subtract the second equation from the first one we have negative to see is negative one, So seize a half, and that implies a Z. I have to. This is the same method over here Let's attract B plus three d equals zero And then we have negative to D equals zero. That means the equal zero. And then that also implies that b A zero. So now we have all four values A, B, C and D plugging those back into our partial fraction to composition. We'LL have one half excellent and then X squared plus three and then one half x x squared plus one on the bottom. So now let's go ahead and evaluate these two. We could pull up those constants, so let's go ahead and do that next. So one half zero one and we see here it looks like you saw will work nicely for both of these. Okay, so just pulled off the constants here. So now for this inner girl, the first integral well, thank you to just be explored blustery then that means to you over two equals x t X, which is up here. So for this integral, we have one half watch out for those limits of an aberration, they may change here you plug in X equals zero and then we give you equals three. That's a new lower bound plug in X equals one and then we get you equals four. So that's our new upper bound. And then we have do you over two on top and then the bottom. We just have a U now for the next in a rule. Same idea. But this time U equals X squared plus one. So do you Over two equals x t x and we see that over here as well. And the numerator and we'll have to watch out for those limits of integration. So now plugging in X equals zero we getyou equals zero squared plus one plugging in X equals one We get u equals two and then up top do you over too. And we have you again. Now it's evaluate each of these Look, we could combine these one halfs We get one over four in a rule of one over you Natural log absolute value you Then we have three to four. So it's for the first thing. Everyone should have kept this in red. So that's the first integral. And then now for the green. No, same integral, just different ballons. So we also have one over for combining those one half actors and then here. We've already integrated natural log, absolute value. You from three to four. So the last thing to do is to just plug in these end points and then simplify. Let's go to the next age for that. So plugging in the four. So through the first, integral on the left, we have Ellen four minus ln three and that for the next one. And lend too minus Ellen ones. We know Elena born a zero. So here we could just go ahead and maybe pull out on one fourth. And then we have Ellen four plus Ellen, too. That's Ellen off four times, too, which is a and then we're subtracting ln three. So that's we could put the three in the denominator inside the log. So really, let me a race that parentheses with Ellen. Eight over three. And that's your final answer.