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JH
Numerade Educator

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Problem 11 Medium Difficulty

Evaluate the integral.

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

Answer

$$2 \ln \frac{3}{2}$$

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Video Transcript

let's evaluate. They give an integral from zero one. So looking at the interim, we see a fraction. So let's see if we can do partial fraction to composition. The first thing that I noticed, it's the quadratic in the denominator, so we should see if this thing factors. Always check to factor the denominator before you do a partial fractions. This the nominating rule factor. We can write it as two X plus one and then X plus one. So this is what the book would call case one. We have distinct linear factors in the denominator, so the partial fraction to composition will be of the form a over two x plus one plus B over X plus one. And then we'LL have to go ahead and find what A and B R and this will be. This Rameau's expression will be easier to integrate. So here, let's go ahead and looking at this equation over here, circled in red. Let's go ahead and multiply both sides by the denominator on the left hand side. So by two x plus one and then X plus one. So let's go ahead and do that and then simplify so on the left hand side will have to. On the right hand side will have a X plus one pleasant B two X plus one and then let's just go ahead and rewrite the right hand side. It's a polynomial. So let's fact throughout the ex term and then we have a and then we have plus to be and then for the constant term with no variable, we have eight times one plus B times one. So on the left hand side, we see that there is no X term or if you want zero X so that means the coefficient of X on the right hand side has to also be zero. So if you want, you could add a zero extra velocity inside that will not change anything. And the coefficients of zero of ex excuse me must match up. And this is why a plus two b has to be zero. Similarly, the constant term on the left is too. The constant term on the right is a plus B. So we have a plus. B equals two. Yeah, and we have a two by two system here involving A and B. Let's go ahead and solve this for and be many ways to do this. If we want, we could take this first equation here, solve it for a so a equals negative to me, and then go ahead and plug this value in for a and the second equation and green. So we have negative to be, which replaces a plus B equals two. So this gives us be equals negative two and then go ahead and solve that for, eh? A equals negative to be. So we have negative two times negative too, which is for Okay, So we've done the partial fraction to composition. We found Andy. So that tells us that we could write the inner girl that's given as the integral from zero to one. No. And then we replaced this fraction with the partial fraction the composition. And then we'LL plug in our values for a baby. So it was for two x plus one and then be was minus two. So if you want minus two over X plus one and these air enrolls you've seen before for the first in a girl, If this two X plus one is bothering you, you can go ahead and do a U substitution not necessary, but it might help you if you're stuck there. And similarly, for the second one, you could also do a use up. U equals X plus one. So let me not do the U substitution here. But you can use these to help you if you're feeling stuck the first integral where? Four times natural log two x plus one. But then I'LL have to divide by two because of this to that pops up over here. So for over two and then for the second integral, I have a minus two and then natural log X plus one an absolute value from zero one so we can go ahead and cancel out that full over to is just too. And at this point, we could just go ahead and plug in our points and then simplify. Let's plug in the one first ssa plugging one. And for X, we have Ellen off two plus one. So that's two natural out of three minus two natural log of two. And then when I plug in zero, we have to natural lot of plugging in zero. We have won minus two natural log of one. We know natural log of one zero. So both of these terms are zero. And then we just simplify this expression over here in the first Prentice's here. You can write this. You can fact rather, too. You have Ellen three minus l in two, and then you could use your log properties to right. This is a log of a Fraction three over, too, and there's a final answer.