5

Find5 _ 9xdx ( = 1) (2 _ 2) where > 1.5 _ 9x| T51)(8r _ 2) dx...

Question

Find5 _ 9xdx ( = 1) (2 _ 2) where > 1.5 _ 9x| T51)(8r _ 2) dx

Find 5 _ 9x dx ( = 1) (2 _ 2) where > 1. 5 _ 9x | T51)(8r _ 2) dx



Answers

Find the indefinite integral.
$$\int \frac{1}{9+5 x^{2}} d x$$

Okay, discussion. We have to solve integration. Two x plus nine d X. And the limit is lower. Limit is five and the limited to Okay, So first of all, we will solve the integration. Then we will apply the limits. So integration of two x dx will be two dot x squared by two. Okay. Plus 90 x. It will be nine x, and the limit is lower. Limit is five and upper limit is two or we will simplify it. It will be X squared plus nine x with limits five and two. So now a parliament to and lower limit five. We will apply limits and the answer. So it will be to Esquire plus 9.2 Okay, minus. It will be five square plus 9.5. Okay. And now it will be to the square well before, plus 9 to 18 and minus five square will be 25 plus nine, five, 45 it will be 20 to minus 70. Or we can say it will be minus 48. So, uh, answer of integration two x plus nine dx with limits 5 to 2. It will be minus 48. And this will be our final answer. Thank you.

Okay, discussion. We have to solve integration. Two x plus nine d X. And the limit is lower. Limit is five and the limited to Okay, So first of all, we will solve the integration. Then we will apply the limits. So integration of two x dx will be two dot x squared by two. Okay. Plus 90 x. It will be nine x, and the limit is lower. Limit is five and upper limit is two or we will simplify it. It will be X squared plus nine x with limits five and two. So now a parliament to and lower limit five. We will apply limits and the answer. So it will be to Esquire plus 9.2 Okay, minus. It will be five square plus 9.5. Okay. And now it will be to the square well before, plus 9 to 18 and minus five square will be 25 plus nine, five, 45 it will be 20 to minus 70. Or we can say it will be minus 48. So, uh, answer of integration two x plus nine dx with limits 5 to 2. It will be minus 48. And this will be our final answer. Thank you.

We are going to be looking at the concept partial fractions. So we have this integral here and we want to take the anti derivative of it. So in order to do that, we're going to have to use the skill partial fractions. So when do we want to use partial fractions? Well, we want to use it when we have a polynomial over another polynomial where the degree of the top polynomial is less than the degree of the bottom polynomial. So in this case, we already have our bottom factored, so it may look like they have the same degree, but in reality, if we were to foil the bottom, we would end up with X squared, which would give us a degree of two vs on the top, we just have X, which means we have a degree of one. So our conditions are met here so we can use partial fractions. Now there is a generic equation for how you deal with partial fractions. So if we have some function p again, that's a polynomial with a degree that is less than the degree of the polynomial cube that's on the bottom. We can rewrite that as a A just representing we literally right in a And we saw for that to find out what that actual numerator is. And that's over. Your first factor. You add that to be again, which will solve for over your second factor et cetera, until you do all of your factors. So we only have two factors here, so we'll only be going to A and b mhm so we can go ahead and write our original in terms of partial fraction. So this is known as the decomposed form. So basically, we're taking this big fraction and were decomposing it into different parts. Now, if your denominator is not already factored and it's fact arable, you're going to want to do that. We're starting with a factor, so we don't need to worry about that just in case you do one or that is not the case. So we have our X minus nine over our X plus five X minus two, and we're going to write that in our decomposed form. So we're gonna have a over our first factor of X plus five. It doesn't matter which factor you do for which one right plus B over our factor of X, my institute. So again, it doesn't matter which factor you do for which, just as long as you do all of your factors. So we're trying to, at least together. And just like when we try to add anything with fractions together, they need to have a common denominator. So in order to do that here, we're going to need to multiply by the other terms, denominator. So for our left term, we're gonna need to multiply by X minus two for both the top and bottom. And for our right term, we're gonna need to multiply by X plus five over X plus five. So again, both the top and bottom so this will get it into our common denominator. So that will give us a Times X minus two plus B X plus five. And then that's all over our common denominator now of X plus five times X plus two. So now we can go ahead and solve. Start looking at solving for what A and B are. So we know our numerator is X minus nine and now we have these denominators being the same. So now we can go ahead and directly set our original denominator of X minus nine equal to our new denominator of a X minus two plus B. Thanks, plus five. So, first, whatever we're trying to solve for things we need to make sure we get rid of our parentheses. So we're gonna need to distribute are a double terms here in r V both terms here. So if we simplify that, we're gonna have X minus nine is equal to X minus two A plus BX plus side and B. So now when we whatever, we're dealing with terms that are not like, like terms, we can only add things that are like terms together. So we know that X is only going to be able to equal terms with exes so we can set those equal in their own equation. So we will have X is equal to X plus BS. Now, since these all have an X in the term, we can actually go ahead and divide each term by X and cancel those out. So we're going to end up with one? Yeah, is equal to a plus B. So that is one of our equations. Now, though, we need to look at a constant. So our constant of negative nine is only going to be able to equal be equal to negative two a plus five B so we can go ahead and write out that equation so we'll have Negative nine is equal to negative to a plus five people. So now we have two unknowns A and B, and we have to linear equations. So now we can go ahead and sub, uh, solve for one of the variable substituted and then solve for that variable, just like we normally do when we have unknowns, multiple equations and we're trying to solve. Now you could use matrices to solve these. If you have a partial fraction that has, say, four or more terms, meaning you're going to end up with four or more equations, you may want to use major cities in order to solve this. Um, it will probably save you time. You can also enter matrices into your T I 83 or 84 calculator if you have one of those. So again, if you have a larger partial fraction, I would recommend using major seats. But we only have two equations here, so we can just go ahead and solve that by hand, so the one equals a plus. The equation is definitely the easier one. So we're gonna go ahead and subtract the both sides I'm gonna solve for a although it doesn't matter which one you saw for as long as you substituted correctly. So we're going to end up with one minus. B is equal to a and we're going to plug this in for our A in this equation here so that we then only have these and can go ahead and solve through the So we're gonna have minus nine is equal to minus two times one minus beef plus five b. Tom. So we're going to need to again distribute this negative two to both terms in these parentheses so we can drop the parentheses and simplify. So we're gonna have negative nine is equal to negative two times one so negative to negative, two times negative B. So it's going to become a positive to be, since those negatives will cancel out and then plus are 5 ft. So now we want to go ahead and solve for B. We already have all our variables and be on the right side, so we're good there. So now we just want to add are to to both sides and already at all the constants on the same side. So negative nine plus two is going to give us negative seven. The tour is canceled here, and then we can go ahead and simplify to B Plus five B, which will give us 7 ft. Now we want to go ahead and divide by seven in order to isolate B and we end up with negative one is people to be so that is one of our solutions. So now we want to just go ahead and plug negative one and four b into this equation, you can plug it into either equation. You can plug it in also to this version or that version or our original over here. Either way, we'll get the same answer for a but we're gonna plug it into this one because it's the easiest. So we have one minus B, which is negative. One is equal to a so minus thing and negative cancels and becomes a positive. So we really have one plus one, which means two is equal to a So now we have solved for both of our numerator so we can go ahead now and put those in. Two are decomposed partial fractions that we had up here. So we're going to have two over our first fraction over our first factor of X plus five. And it's going to be minus one since it was negative one and brought the negative out in front X minus two. So now we have our original decomposed into different terms. Now we can go ahead and go back to taking are integral, which was the point of why we're doing this. So now we have two separate terms, which means we can take the integral of each term separately. Now, the key to taking the integral of these of to over expose five and negative one over X minus two is we need to know what the anti derivative of the of one over X is some kind of our generic function. So the anti derivative of one over X is Ellen of the absolute value of it. So we're going to apply that principle over here. So too is just a coefficient. So to would be out here then the anti derivative expose five is going to be the Ln of the absolute value of exposed five. Make sure you do the absolute value that is important as otherwise. You could end up with a negative there and you can't take the look of a negative. It's not in its domain, so it's important to put that absolute value there. Then we have negative one as a coefficient. Since we're multiplying by one, we don't need to actually write the one what we need to make sure we include our negative. So then the anti grid of X minus two is going to be the Ln of the absolute value of X minus two. And then, since we are doing an indefinite integral, we're going to need to add our plus C that we must always add whenever we are doing indefinite integral. So this would be our final answer here. So once you get through decomposing it and getting it into its different terms, taking the integral should not be too difficult. And that's part of why we put all of this work into the beginning so that then the end is pretty easy Now. Note. If we had coefficients on these exes here or had X squared or something, we would need to take the chain rule into account. So make sure that you look at that. Here are coefficients were just one, so we didn't need to worry about that. And then another note is just if you have a factor that appears more than once here, both these factors only appeared twice. You do write out your decomposed form a little differently. So if you have a repeated linear factor, you would need to have a partial fraction for each power of that factor up to and including the NFA power. So if we had similar to what we started with, like negative nine over X plus five squared and then X minus two, we would write that a little differently. The numerator parts still gonna be the same a b c d e etcetera. But so we have our first factor. So X plus five. But then for our second factor so far, plus B, but we're going to write it as X plus five squared. So the 10th term of our X plus five is too. So we're gonna write one for just X plus five and then for X Plus five squared. And if this was X plus five Cube, we would go ahead and write another one. So plus the over X plus five cubes. So all the way up until you get to whatever the end term is, And then if you have any other factors, you would just add them like you would normally. So we do, plus c X minus two. So that's how we would write out that one if we were trying to do, um, partial fractions on that. So that is how you deal with partial fractions in general. And I hope this was helpful, Yeah.

Again Discussion. We have to solve integration. Two X plus nine b X, and the limit is lower. Limit is two and upper limit is five. Okay, so to solve this, first of all, we will solve the integration part. And then we will have, like, the limits. So integration of two x dx, it will be too X squared by two. Okay, plus integration of 90 x will be nine x. Okay. And the limited total five. Or we simplify it. It will be X square plus nine x, and the limit is 2 to 5. Now, we will apply the limits here, so it will be a parliament minus lower limit. So x square, it will be five square plus 9.5. Okay. Minus X square for lower limit. It will be too square, plus nine dot x. It will be 9.2. So it will be 25 plus 9, +495 45 minus. It will be four plus 18. Or we can say 25 plus 45. That will be 70 minus four plus 18. That will be 22. And it will be 48. Okay. Or we can say the integration of two X plus nine DX with limits 2 to 5. The answer will be 48. Okay. And this will be our final answer. Thank you.


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