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IntegrateJ (4r' 92 + 13) d...

Question

IntegrateJ (4r' 92 + 13) d

Integrate J (4r' 92 + 13) d



Answers

Calculate. $$\int \frac{d x}{x^{2}-4 x+13}$$

We have the integral of three I plus four t J D t. So we're going to take the integral of each component. That's three t plus C one, and the integral of the second opponent component is to t squared plus C two. Notice that I'm labeling the components. Excuse me. I'm labeling C one and C two because it's possible that the values of C one and C two are different. Okay, and this is it.

Neither In this problem, we're supposed to compute the definite integral from 0 to 3 of the norm or the length of this vector here. T I plus t score, Jay. So at first, this looks like it doesn't match up with any formulas that we learned in this chapter. But let's just remember how to compute the length of the vector. And this will end up looking ah, like something out of college one instead of Coke three. But getting there is a little tough. So if we just rewrite everything, but we know how to compute the norm of the vector, it's a square root of the sum of the squares of its components. So the first component is T. We'll square that to get t squared. The second component is t squared, so we will square that to get t to the fourth. And so that is the norm of our vector That's in here. So we just rewrote that. And now we have just integral on one variable. So, uh, we know how this works anyway. Now, what can we do to simplify notice? We can factor a T squared out of the square root, which oneness with one plus, uh, T squared and Sund. And we're happy about this because now that scored of T square that can come out as just a t. So we're left with t times the square root of one plus t squared. Now, we could use use u substitution here if we wanted where you may be able to do this in your head at this point, Um, I'm just gonna rewrite this as a power. So the 1/2 power and again, if you can use u substitution, or you may just be able to see that this tea is the derivative of one plus t squared, except we're missing a factor of two, so are anti derivative. When we actually compute this using the fundamental theorem are anti derivative. Make sure you're convinced will be 1/2. That's because of the missing to factor in front of the tea and now at the power of 1/2 would go up by one. So we get three halves as usual. We will divide by three have, in other words, 2/3 in front. The one plus two square there, this is our anti derivative will clean cleaning up in a minute. Um and then we'll import three and zero again. If you want to use u sub, that's fine. Just make sure you get the same thing as we do here, and we'll go back over here just to save space. Obviously, obviously the twos cancel out. So we end up with 1/3 just a constant Let's plug in 34 are variable T. So one plus three squared will give us 10 so they tend to the three halves. Now we have to subtract the result of plugging in zero. So one plus zero, of course, is one to get one of three halves. In other words, we get 1/3 of tend to the three have minus one, and that is our final answer. So hopefully this helped

Hello France. We have to find the integration of this given expression that is integration of nine into our to the power minus four d. r. okay yeah so we can die. Non integration of art department is four India. Yeah so we know that integration formula indication of U. To the power and do you. So that will be cause to You. Did the government class Wanna born and plus one press constancy. We will use this formula. So this will be cause to nine into our to the power minus four plus one Upon -4-plus 1 Plus constant C. So this would be because 29 upon monastery R. To the power of monastery Plus constancy. So we can write minus of three R. To the power Monastery plus question. See so far that we can die- of three upon aren't you Plus constancy. So this is that answer I hope you understood. Thank you.

We're going to find that if indefinite integral of the vector valued function shown. So what we're gonna do is we're going to be integrating each of our components separately. So think of a one over T or a D T over T. That's our case of Ln so we'll get the Ellen of the absolute value of T. In the I direction. There was a one in front of J that becomes a. T. In the J. Direction and then we go up a power to five halves and multiply by 2/5 multiply by the reciprocal, same as dividing by the new exponent. And at the very end I say plus C where C is an actually vector valued function. And so um we would use all of our components. They also could have each been written uh in our actual components.


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