Evaluate the integral. ∫ 0 3 3 x + 10 x 2 + 9 x + 20 d x \int_0^3\frac{3x+10}{x^2+9x+20}dx

this problem, we're asked to evaluate the following integral. And here we have the definite integral from zero to three of three x plus 10 over x squared plus nine x plus 20. How do we solve this case? Well, in this problem here, what I can first do is take a look at my integral to see if there's any place where I can do some kind of variable substitution or something that's gonna make this more simple. Now you could try a variable substitution on the denominator here, or you could try it on the numerator. I'll tell you right now, though, if you went through that process, you would find that as you took the derivative of either the numerator or the denominator or whatever you said equal to u, it would not make the integral any more simple. So what I need to do here is I'm going to need to find some other strategy to simplify this rational rational function to a way where I can solve it. Now what comes to mind is using partial fraction decomposition. Now first off, when using partial fraction decomposition, I need to take a look to see the top degree or the highest degree in the numerator versus the denominator. I can see that we have a degree of one in the numerator and a degree of two in the denominator. Because of that, that means that I will not have to do polynomial division as a first step. So what I need to do is find some sort of way to factor the denominator at this point. So let's see how we if we can figure this out. This is the integral from zero to three, and that's going to be three x plus 10 on the top. And then on the bottom, well, I need to think about a pair of numbers that is going to multiply to give me 20 and add to give me nine. What two numbers would that be? Well, thinking through this, five and four seem to be good choices, because five times four is 25 plus four is nine. So let's break this denominator down into x plus five times x plus four. Now notice at this point, I have this broken down into distinct linear factors. So this is a case where I could perform partial fraction decomposition. Now what I'm gonna do is I'm gonna take this portion of the fraction right here. I'm gonna specifically focus on this for now. So we're going to ignore the integral for the moment here just because I wanna get this fraction broken down. So let's create a separate space down here where I go ahead and try to break this fraction down, three x plus 10, this rational function, over x plus five times x plus four. And let's see how we can break this down into separate multiple simpler fractions. While doing this, I know since we have a distinct linear factor right there, I'll have some constant a over x plus five. And since I have another distinct factor here, I'm going to have some constant b over x plus four. So that's what happens when we break this up. Now from here, my next step is going to be to take everything in this equation and multiply it by the common denominator. In this case, I can see the common denominator is x plus five times x plus four. I need to go ahead and distribute this into both sides of the equal sign. So how can I go about doing that? Well, first off, if I go ahead and distribute this common denominator right here, notice that it's going to go ahead and cancel completely. So because of that, all I'm gonna be left with on the left side here is three x plus 10. Now what about right here? Well, if I multiply this a term by the common denominator, notice the x plus five cancels entirely, so all I'm going to be left with is a times x plus four. Now what about this b term? Well, if I multiply this by the common denominator, notice the x plus four will cancel completely. So that's just going to give me b times x plus five. So now I have this equation. Now from here, what I want to do is I want to think about how I can go about solving for my constants a and b. Now there are some strategies I know to solve this. I could try expanding everything, grouping, and getting a system of equations, or I could do strategic substitutions for x. Since I can see there are really only two factors I need to worry about here, I think if I just make a couple of strategic substitutions, that will allow me to get my constants pretty quickly. So first what I'm going to do is let x equal negative four. Doing this will give me negative four plus four, which will allow me to completely remove this factor. So that's going to give us three times negative four plus 10 is equal to a times negative four plus four plus b times negative four plus five. Now negative four times three is negative 12 plus 10. That's gonna give me negative two. That's what this whole thing comes down to. Then we'll have that equals a times negative four plus four, which I know this whole thing is just going to equal zero. So all I'm going to have is b times negative four plus five, which is the same thing as b times one, or just b. So we have that b is equal to negative two. So going up to this fraction up here, we can say that our b is negative two that we're dealing with, meaning it's also going to be negative two down here. So that's how we can go ahead and get our b. Now next I need to try and find my a. So what can I do here? Well, if I'm looking to find a, I'm going to go ahead and choose another value for x. In this case, I'm going to let x equal negative five. So I know negative five plus five is just going to give me zero. So in this case, we are going to have three times negative five plus 10 is equal to a times negative five plus four minus two times negative five plus five. Now in this case, the negative five plus five is zero, so that whole term is just going to go away. So we're going to be left with three times negative five, which is negative 15, plus 10, which is negative five. That's gonna be equal to a times negative five plus four, which comes out to just negative a. Canceling the negative signs will get that a is equal to five. So I can say that our a up here is five. So we've now found our a, and we've found our b. So in this case, we've gone ahead and successfully performed this partial fraction decomposition. So going back up to the integral we had right here, this integral is equivalent to the integral from zero to three of five over x plus five, minus two over x plus four, all integrated with respect to x. Now at this point, this is an integral that I can do, because what I can do is integrate each of these terms separately. The integral of five over x plus five, well, that five would come outside the integral, so we'd just be integrating one over x plus five, which would be five times, and then the integral of this would just be the natural log of x plus five. Don't forget these absolute value bars here. Then we'll have minus, and then I need to go ahead and integrate this portion. Well, that constant is gonna come outside, so I have minus two. And then integrate this, we'll get the natural log of x plus four. And then keep in mind this whole thing is going to be bounded from zero to three. So next, I would want to apply these bounds. Now I can go ahead and do this below. So I'm gonna go ahead and make some space up here, or I'm just gonna move this down here so we have some more space to solve it. So this is what we're dealing with right here, and what I can go ahead and do is I can apply these bounds now. So applying these bounds to this this whole expression we have right here, Well, first, I'm going to plug in my high bound of three. So we're going to get five times the natural log of three plus five, which is going to be eight, so the natural log of eight, minus two times the natural log of three plus four, which is seven. So that's what this whole thing is going to come out to, and then we're going to have minus, then I'll plug in zero. So we're going to have minus, then we're going to have five times the natural log of zero plus five, which is five, minus two times the natural log of zero plus four, which would just be four. Now all of this is something you can type into your calculator right about here, and I encourage you to do that and to evaluate this whole expression. And what this all should come out to, if you type this in to your calculator, you should get that this whole thing is approximately equal to, as a decimal approximation, 1.23 approximately. So about 1.23 is going to be this value as an approximate decimal. So going up to our integral right up here, we can see that this whole thing comes out to approximately 1.23. So about 1.23 is gonna be the solution to this problem, and that is how you can deal with these types of definite integrals. So as you can see, it did take a lot of steps, but the main thing was to go ahead and do this partial fraction decomposition right here. Once we did that, we went ahead and were able to split this into two fractions and integrate each one separately and apply the bounds at the end. So hope you found this video helpful. Let me know if you have any questions, and let's move on.

12. Techniques of Integration

Partial Fractions

Calculus

12. Techniques of Integration

Partial Fractions

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Multiple Choice

Evaluate the integral.
$\int_{0}^{3} \frac{3 x + 10}{x^{2} + 9 x + 20} d x$

$6.51$

$negative 13.22$

$negative 4.19$

$1.23$

Verified step by step guidance

Step 1: Recognize that the integral involves a rational function, $ \int_0^3 \frac{3x+10}{x^2+9x+20} \, dx $. The denominator $ x^2 + 9x + 20 $ is a quadratic expression, so we should first factorize it.

Step 2: Factorize the quadratic expression $ x^2 + 9x + 20 $. This can be written as $ (x+4)(x+5) $, since $ 4 \cdot 5 = 20 $ and $ 4 + 5 = 9 $. Rewrite the integral as $ \int_0^3 \frac{3x+10}{(x+4)(x+5)} \, dx $.

Step 3: Use partial fraction decomposition to break $ \frac{3x+10}{(x+4)(x+5)} $ into simpler fractions. Assume $ \frac{3x+10}{(x+4)(x+5)} = \frac{A}{x+4} + \frac{B}{x+5} $, where $ A $ and $ B $ are constants to be determined.

Step 4: Solve for $ A $ and $ B $ by multiplying through by $ (x+4)(x+5) $ and equating coefficients. This gives $ 3x+10 = A(x+5) + B(x+4) $. Expand and collect terms to find $ A $ and $ B $.

Step 5: Substitute the partial fractions back into the integral, $ \int_0^3 \frac{A}{x+4} \, dx + \int_0^3 \frac{B}{x+5} \, dx $. Evaluate each term using the natural logarithm formula $ \int \frac{1}{x+c} \, dx = \ln|x+c| + C $, and apply the limits of integration from 0 to 3.