Partial Fractions Example 3 Hi everyone! Welcome back. We’re going to do another partial fractions problem today. This one is the integral of dx/x2+x. The first thing I’m going to do is change this integral a little bit so that it’s simpler. I’m going to take the dx out and then factor d denominator. So this is going to end up as 1/dx, I took the dx out and then I’m going to factor out an x in the bottom, so x(x+1). I’ve simplified the problem and it becomes apparent now that this is a candidate for partial fractions problem because we have two terms in the denominator. One is x and one is x+1. With partial fractions again, the first thing we do is write 1/x(x+1) on one side, we set that equal to A over the first term in the denominator x plus B over the second term in the denominator, x+1. And we do that so that we can solve this equation for A and B. The first thing we do, we multiply each of these terms by x and x+1. So we can see x(x+1), we can see here that if we do that everything over here is going to cancel except for one, so one equal. X is going to cancel here, so we’ll have be left with A(x+1) plus, x+1 will cancel here so we’re left with B(x). You can see if you want to skip this step here, you can just take the numerator 1 until that equal to A times one of the terms here in the denominator, plus B times the other term. It doesn’t matter what’s multiplied with A and what’s multiplied B, you could flip this as long as you keep your A and B consistent. So you can skip this whole step and just go straight to this and make it easier on yourself. We end up with this equation here and what we need to do is solve for A and B. The way that we do that is by canceling out A first and solving for B and then canceling it out B and solving for A. So first let’s cancel out A, the way that we do that is set this right here equal to 0. If we put -1 in for x, we’ll have -1+1 and that equal 0. And if this is 0, then you’re A(0) and A cancels out because it’s multiplied by 0. So we’re going to go ahead and set everything in the equation equal to -1 so that A will cancel and we’re going to have 1 equals, remember A cancels because we plugged in -1 and this is at being 0 and we’re left with B(-1). Therefore, 1÷-1 is -1 and we get the equals -1. So we solved for B. Now we need to solve for A and the way we do that is by canceling out B. If we put 0 in for x then B will cancel, B(0) is 0. Let’s go ahead and put 0 in for x, we’ll have 1 equals A times, and we said we’re going to put 0 in. So 0+1 is just 1 and then of course this is going to be 0 do we can just leave that out. We can see that A=1. So we solved for A and we solved for B. Now what we need to do is plug A and B back into this part of our equation here and then we plug that whole thing back into the integral. We’re going to have the integral of A which is 1 over x plus B which is -1 over x+1. This is something that we can now take the integral of. We solve for A and B, we plug them back into here and then we put this whole thing back into our integral. It’s just basically a way of transforming the original problem into something that we can actually take the integral of and the way that we do that is partial fractions. So now that we’ve got this, the formula that we need to use is this. The integral of 1/x equals ln or natural log of the absolute value of x. This is a formula, it’s a rule. You need to know what for partial fractions. Almost partial fractions problem at least the basic ones and we’re going to use this rule. You can see we’ve got 1/x here, the integral of that is natural log of the absolute value or taking one term at a time. So we’re going to do this one first and then this one. So the integral of this 1/x is natural log of the absolute value of x and since it’s a plus or a negative. We’re going to do minus natural log of the absolute value of x+1. So the denominator always gets inserted into this absolute value brackets this ln and don’t forget to add C to ac