Finding the integral of a rational function using linear partial fraction decomposition.
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- I think that at the end of the vedio, the answer of that question should be -7In|2x-3|+4In|x-1| instead of -7/2|2x-3|+4In|x-1| because the numerator's 2 can be cancel out. Can anyone help me with that?(2 votes)
- Sure, it's because of the chain rule. Remember that the derivative of 2x-3 is 2, thus to take the integral of 1/(2x-3), we must include a factor of 1/2 outside the integral so that the inside becomes 2/(2x-3), which has an antiderivative of ln(2x+3). Again, this is because the derivative of ln(2x+3) is 1/(2x-3) multiplied by 2 due to the chain rule.(9 votes)
- 6:10Hi isn't 2 over anything to the minus one integrated, not 2ln? 2ln multiplied by -7/2 is not -7/2ln it would be -7ln? I'm so confused, have I been doing it wrong all this time.
Edit: I have become aware of my mistake I was doing a lot of it in my head and I forgot the du= 2dx would remove the 2 on top of the ln making it 1/(x-3) and so it would be -7/2 multiplied by 1ln and not 2ln. I'm leaving my comment here in hopes that it will help those who have made the same mistake.(4 votes)
- Can someone point me to a video on partial fractions? The step from x-5/((2x-3)(x-1)) to A/(2x-3) + B/(x-1) was not entirely clear to me.(2 votes)
It looks like partial fraction expansion is the reverse of the method used to add rational expressions from algebra.(3 votes)
- Why does Sal write the answer in terms of the absolute value of the expression? Why not just ln(2x-3), etc?(3 votes)
- I think I've spotted a little mistake at6:30: When you simplify the fraction of -7/(2x-3), you take out -7/2.
Shouldn't the fraction become 1/(x-3) instead of 2/(2x-3) or am I missing a step in this calculation?(1 vote)
- You're missing a step.
The reason it becomes 2 is because you are putting the derivative of (2x-3) as the numerator.(3 votes)
- Can't you simply simplify and use u-substitution?(1 vote)
- What do you mean exactly? you could, for example, divide the expression and get the "quotient plus remainder over denominator" form. In fact, the quotient is quite easy to integrate (not even u-substitution is required) but integrating the remainder divided by the denominator requires you to integrate a product of two function...to the minus one power! Unfortunately, i'm not able to do that and it's at least as complex as the problem we started with. Hope that this makes sense(2 votes)
- a question from part of the practice problem:
- They aren't the same. ln|2x-2| equals ln(2) + ln|x-1|. So, there's an additional ln(2) term there.
Considering this is integration, you could condense the ln(2) into the constant of integration at the end. So, ln|x-1| + c and ln|2x-2| + c would be the same (though the value of c in the first expression would actually be (ln(2) + c)). However, in general, the expressions aren't the same.(2 votes)
- I think there is a mistake around4:10. Shouldn’t the equation be A + 3B = -5? Because you have x-5.(0 votes)
- Yes, however, (A+3B) is initially negative, or being subtracted from (A+2B)x, and so Sal skips a step and multiplies both by -1, resulting in (A+3B)=5. Does that help?(3 votes)
What if the denominator is not factorizable??(1 vote)
- probably have to find a way to factor it even if the factors are irrational, and then you can do the same process from there(1 vote)
- How would it be different if there was a number as the numerator with no x included?(1 vote)
- The process would be the same, we would just force A(x-1)+B(2x-3) to equal that number instead of the polynomial.
If the number is 1, we would need A(x-1)+B(2x-3)=1 for all x, so we would require Ax+2Bx=0 for all x (so A+2B=0) and
-A-3B=1. Solve the system to get B=-1, A=2.
So the rational function would split as 2/(2x-3) - 1/(x-1).(1 vote)
- [Instructor] We are asked to find the value of this indefinite integral. And some of you, in attempting this, might try to say, all right, is the numerator here the derivative or a constant multiple of the derivative of the denominator? In which case, u-substitution might apply, but it's not the case here. So what do we do? And my hint to you would be partial fraction decomposition, which might invoke some memories from a precalculus class or maybe from an algebra 2 class, but it's a technique to break up this rational expression into the sum of two rational expressions. And a good hint there is the fact that this denominator here is factorable into these two expressions. So what we're gonna try to do with partial fraction decomposition is say, can we express x-5 over (2x-3)(x-1), can we express it as a sum of two rational expressions, where the denominator of the first rational expression is 2x-3, and the denominator of the second rational expression is x-1, and I don't have to put parentheses there, is x-1. Now we don't know the numerators. As you might have learned before, and I encourage you, if this the first time you're ever seeing partial fraction decomposition, look that up on Khan Academy. We have many videos on it. But the general principle here is that your numerator is gonna be one degree less than your denominator. So our denominators here are first degree, so our numerator's gonna be zero degree, or just constants. So it's gonna be some unknown constant data. Let's call that A. And some unknown constant B. And our goal is to solve for A and B. And this is all a review. This isn't really calculus. This is more precalculus now, or algebra. How do we solve for A and B? Well, let's add them as if we're just adding two fractions with unlike denominators. So we wanna have a common denominator. So what we wanna do is multiply this first rational expression by x-1 in the numerator and the denominator. So you could call this A(x-1) over (2x-3)(x-1). And then for this second rational expression, we'd multiply the numerator and the denominator by 2x-3. So (2x-3)B over (2x-3)(x-1). And now, since I have the same denominator, I can add them. And the goal is when I add 'em, and I think about the numerator, I will try to say, well how do the A's and B's line up to what we have already, right over here? So this is going to be equal to, same denominator, so the denominator's (2x-3)(x-1), and then in the numerators, actually, let me do it over here. This is the same thing as Ax-A, just distributing the A, and if I distribute the B here, this is the same thing as 2Bx-3B. And so if we add the numerators, we can add the x terms, Ax+2Bx, so we could call that (A+2B)x. And then if we add these constant terms, -A, -3B. And since we have a minus right over here, I'm gonna try to pattern match, so I'm gonna have minus, and I'll write that as -(A+3B). If you distribute this negative sign, then you're gonna get -A and -3B. So these are equivalent. And now we can see the pattern. We say, all right, our denominators are now the same, so this thing in our numerator has to be the same thing as x-5. So whatever our coefficient here is on x, that's gotta be equal to 1, that's the coefficient on x. And this thing that we're subtracting, that's gotta be equal to 5. And we're gonna set up a system of two equations with two unknowns to solve for A and B. So we know that A+, so let me write that down, we know that A+2B needs to be equal to 1. And we know that A+3B, or A+3B, is going to be equal to 5. And so now, to solve for A and B, well, we could do that by elimination. So let's see, what if we multiply this top equation by -1. So that'd be -A, -2B, -1, and now we add them together. The whole point was to cancel out the A's, so we're left with -2B+3B, that's just going to be B, is equal to 4, and then we can substitute back in to solve for A. So let's say right over here, we know that, we know that A+3x4, so that's gonna be 12, I'm using this equation right over here, is going to be equal to 5. Subtract 12 from both sides, you get A is equal to -7. So just like that, we can rewrite this entire integral. We can say this is going to be equal to the indefinite integral of, open parentheses, A over 2x-3. We now know that A is -7, so it's -7 over 2x-3, and then we're going have +B, B is 4, so, +4 over x-1, over x-1, and close parentheses, dx. Now if you are so inspired, I encourage you to pause the video and try to run with it from this point, because we have seen the techniques to solve integrals like this before, but I'll do it step by step. This is going to be the same thing as, well actually let me just, so this is going to be the same thing as the integral of, so 2x-3. I could write the -7 here, but I'm gonna take the constant out of the integral. So I'll put a -7 here. And to help us solve this, and this could be a 1, but to help us solve this, it would be nice if we had a 2 here. Why is that? Because 2 is the derivative of 2x-3, and so then we can do our u-substitution, which we have, sometimes, gotten practiced doing a little bit in our head. And so if we want this to be a 2, we can't just multiply by 2. We've also gotta divide by 2. And I can do that outside, 'cause that's just a constant dx+, well the derivative of x-1 is 1, so we just want a 1 up in the numerator. So we can take the 4 out of the integral. 4 times the integral of 1 over x-1, dx. So this is going to be equal to, we just have our constant out front, -7/2, and since we have this thing in the denominator, and we have its derivative, we can really just think about this as integrating with respect to this thing in the denominator. You can sometimes view u-substitution, which I'm not explicitly going to do here, as the reverse chain rule. The antiderivative of 1 over x is the natural log of the absolute value of x, but here, this is going, the antiderivative of this is going to be the natural log of the absolute value of 2x-3. And then for this part, it's gonna be +4, times the antiderivative here is the natural log of the absolute value of x-1. Once again, I can do that, 'cause the derivative of x-1 is just 1. And of course, since we're taking an indefinite integral, we do not want to forget our +C. And we're done.