5

QUESTION 25Find dx Nx+ 3 + Vxa No correct answer ; (513)| (x+ 3)} b_ ~X +C; (10/9/(x+3)} +x C} (10/9)| (x+ 3)} -X ce.9Nx-34*+3+C...

Question

QUESTION 25Find dx Nx+ 3 + Vxa No correct answer ; (513)| (x+ 3)} b_ ~X +C; (10/9/(x+3)} +x C} (10/9)| (x+ 3)} -X ce.9Nx-34*+3+C

QUESTION 2 5 Find dx Nx+ 3 + Vx a No correct answer ; (513)| (x+ 3)} b_ ~X +C ; (10/9/(x+3)} +x C } (10/9)| (x+ 3)} -X c e.9Nx-34*+3+C



Answers

Solve the given problems. Explain your answers. Is $\int x^{-2} d x=-\frac{1}{3} x^{-3}+C ?$

Hello. We have to explain our answer that integration is given to resolve 2. 1. Holy Square The X- 80 plus two to Express 1 to the power of three plus C. Is this correct or not? So We will take a left inside integration of three strikes off to explosive one holy score of DX. And we will take to express one urgent calls to one. So this will be by differentiating it with distract well so this will be cost twice over the years because to do you so we can try off you square into the way two. Okay So this will be cost 23 x two of years square. Do you know this will because three x two of you To the power of three Upon three plus concern see so this will be a cost to one way two of you that is to Express 1 to the power of three this question see so this will be the answered. Okay There should be a factor of one x 2 in the result. You can see that in the given result in the question there is no factor. There should be if you're traveling by do for the characters are Okay, so we can do I know this is not the character isn't this one is the correct result, so we cannot know. There should be a factor of one way to it's Hector off one between the result. As you can see that from he had this factor is one way to I hope you initial Thank you.

Given a u of expose three. And given the fact that do you is detox and you minus three is acts we can first substitute to get the integral zero minus three. You want half times, do you? This gives us once we've integrated 2/5 you to the five over two minus three temps. 2/3 you the three over two plus c, which is equivalent to So we're substituting back ends. We're getting rid of you more. Substituting back in with the actual value is minus two times X plus three the three over two plus C Sansa trice bee.

The institution we have to solve the double integration that is integration from 0 to 1. And integration from X two row text X squared plus Y square and two dy Dx. So this can be written as integration 0 to 1. And I am going to integrate with respect to Y. So this will come out to be excess square, Y plus Y que by three. And the limit is from X to wrote checks into dX. Now put this limiting the This integration so this will come out to the integration 0-1 and this is X to the power five by two plus one by three. X square. Sorry one x 3. This is extra the poetry by two. Now I'm going to put this limit so this will come out with minus X cubed minus excuse bigotry into the X. Now I'm going to integrate it with respect to X and I will get disease two by seven plus one by three and two To buy five one x 4 -1 x two. And after solving this, I will get disease three x 35. So this is our answer for this constitution. And for that option, B is the correct choice. Thank you.

Welcome to this lesson. So in this lesson will be next to verify whether or not one over X plus five all to the path three is the correct anti derivative of one over three out the next plus five all to the ball too. So here mostly when you have the power of the function, it's a positive power at the numerator side, it would need to come up. Okay, so it comes up like 1/3 then X plus five towards the bar negative too. So because it's down and the negative comes here, so that in this time when you subtract one from it it becomes A -3. Okay, so the negative story comes down and now when you're multiplying it by the as permanent the power it leaves a negative trees out there. Okay. So the reason why it has become the negative story here, uh the three there is that it has left a negative sign over there. So this is not the anti directive. The anti directive. This is not the original function. Original function is rather negative. Fool there. Yeah. Okay. Yeah. All right. Okay. So thanks for a time. This is the end of the lesson.


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