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Evaluate the integral.

$ \displaystyle \int_0^1 \frac{x - 4}{x^2 - 5x + 6}\ dx $

$$

\int_{0}^{1} \frac{x-4}{x^{2}-5 x+6} d x=\ln 3-3 \ln 2=\ln \frac{3}{8}

$$

Integration Techniques

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Missouri State University

Harvey Mudd College

University of Nottingham

let's use partial fraction to composition to evaluate the definite integral from zero one. Looking at this fraction, we see that there's a quadratic in the denominator. So let's go ahead and try to factor that. So does this denominator factor. Sure does weaken. Take X minus two X minus three. So this is what the book would call case one. We have a distinct linear factors non repeated. So we have hey, over X minus two plus B over X minus tree. So this is the form for the partial fraction to composition, and I will have to use some algebra here to sell for Andy. So looking at this equation here circled in red, that's good and multiply both sides by the denominator on the list. So multiplying by X minus two times X minus three. On the left, that denominators will will cancel. And on the right hand side, we have a minus three plus B X minus two. Let's just go ahead and rewrite the right hand side. Let's pull out the ex term. We have X and then we have a Plan B and then we have minus three, eh? Minus two b. It's on the left hand side. We see that the coefficient in front of the ex is the one on the right. It's a plus, B. So we must have a plus. B equals one. Similarly, the constant term on the right is on the left. Excuse me is minus four, but on the right hand side, the constant storm is minus three minus two B. So this gives us a two by two system of equations too soft for A and B in that last equation there, you could go out and multiply by a negative one many ways to solve these systems. For example, we could take the first equation, solve it for Bea, and then we could plug this value of be into the other equation and in software, eh? And doing so will have three a plus two, one minus a equals four. And this will give us a plus two equals four so equals two and then must take this value of a plug it into this equation up here and we get B equals one minus two, which is minus one. So we found an B. So now we're ready to integrate. Let's come to the original problem the original integral. In the top left, we can rewrite this. So we replaced the fraction with the partial fraction to composition over here in the far right upper right corner. And we've just found A and B. So I have two over X minus two and then for B, we have a minus one over X minus three, and this new integral should be easier than the original. Now, if this X minus two or explain histories bothering you, you can go ahead and do a U substitution. U equals X minus two and for the second one, x minus three. And by doing so after simplifying, we should have to natural log absolute value X minus two minus natural log absolute value X minus three from zero one. Let me separate this from our scratch work. We've evaluated the inner girl. We have the natural logs. Now we just plug in the end points and simplify. So we plug in one first. We have natural log of negative one, but we have absolute value, so that just becomes a one. Then we have natural log of one minus three is negative to take the absolute value you just get into there and then when we plug in zero, the first term will become two times natural log. We have a negative to their absolute value. Makes it positive. And then finally, we'LL have a natural aga positive three. After we take absolute value and then just go ahead and simplify as much as you can. We know natural Aga one zero Combining these national log of twos. We'LL have negative three Ellen, too, and then watch out for this double minus here. That becomes a plus at one three. And there's many ways to simplify this. For example, we can write. This is Ellen three minus ln eight. By taking that using one of the log properties that says you can take the coefficient in front of the log, bring it inside. But then you have to raise it to the exponents. So two to the three is eight, and then I could use another property of logs log. If you subtract the lines, you can rewrite it as a fraction log of a fraction. So it looks like I have any one of these will be the right answer. Let's just maybe narrow down some of these two right here, and that's your final answer