Question

Show that

∫√(2+x^2)dx = x√(2+x^2)/2 + ln(x+√(2+x^2)) + C

where C is an arbitrary constant

I don´t even know where to start :)

Answer 1

tan or sec sub

Answer 2

oh we are just showing, just take the derivative of the right side

Answer 3

May I help?

Answer 4

@JohnW are you there?

Answer 5

seems not

Answer 6

Yes

Answer 7

Thank you.

Answer 8

Hope that helps

Answer 9

I´m sorry, but I don´t follow. I don´t even know what an arbitrary constant is :/

Answer 10

Even less knowing how to show that C is one

Answer 11

You are just starting derivatives?

Answer 12

I think they just want you to derive the right side and show this statement is true

Answer 13

the derivative of a constant is 0, as a derivative tells you the rate of change wrt some variable, and a constant has no change

Answer 14

first thing you should be aware of is that derivative is a linear operation

Answer 15

so that means
d/dx(x√(2+x^2)/2 + ln(x+√(2+x^2)) + C)
=d/dx [x√(2+x^2)/2] + d/dx [ ln(x+√(2+x^2)) ]+ d/dx[ C ]

Answer 16

work on deriving one term at a time, mar3 will help u i have to help someone rn

Answer 17

ok, thanks!

Answer 18

@JohnW

Answer 19

Like what 518nad said:

Derive the right side and compare to the left side, if identical, you are correct, if not you may need help

Answer 20

\[\frac{ d }{ dx }(\frac{ x \sqrt{x^2+2} }{ 2 })=(\frac{ x }{ 2 })(\frac{ 1 }{ 2 })\frac{ 2x }{ \sqrt{x^2+2} }+\sqrt{x^2+2}(\frac{ 1 }{ 2 })\]

Answer 21

\[\frac{ d }{ dx }(\ln(x+\sqrt{x^2+2}))=\frac{ 1 }{ x+\sqrt{x^2+2} }(1+\frac{ 1 }{ 2 }\frac{ 2x }{ \sqrt{x^2+2} })\]

Answer 22

\[\frac{ d }{ dx }(C)=0\]

Answer 23

Sum all the three results, and simplify the terms

Answer 24

ok

Answer 25

Do you want to do it for you or you can do it?

Answer 26

@JohnW
Let me know please, brother!

Answer 27

Hope that helps

Answer 28

Please help me :)

Answer 29

I´ll only get some of the numbers mixed up and then the entire calculation is screwed up ;)

Answer 30

Where are you stuck?

Answer 31

Brother!
Where are you stuck?

Answer 32

I`m sorry. I have a lot going on at the same time. I think I can follow the process, but I´m not familiar with how you simplify the result.

Answer 33

I am busy too
but when you are ready to continue, please let me know.

Answer 34

Thanks. now I`m online again and I`m looking for help with this question.

Answer 35

Sorry, I was not here!

Answer 36

Thank you for the medal!