Question

calc!

Answer 1

@Michele_Laino

Answer 2

there are two methods in order to compute your integral

Answer 3

please try this substitution:
\[t = \sinh \theta \]

Answer 4

okay(:

Answer 5

since it is squared, it can come out of the sqrt

Answer 6

you have to apply this identity:
\[{\left( {\cosh \theta } \right)^2} - {\left( {\sinh \theta } \right)^2} = 1\]

Answer 7

ohh got it

Answer 8

please note that if you are not expertise with hyperbolic functions, then you can try another method

Answer 9

i think i got it(: i got my answer

Answer 10

ok!

Answer 11

let me know if im totally wrong... 1/sqrt2 - x/ sqrt 1+x^2

Answer 12

Is your final result?

Answer 13

yes..

Answer 14

Please wait I check your result

Answer 15

ok(:

Answer 16

no, I got a different result, please retry and write step by step

Answer 17

wait was my first answer correct then? sqrt 1+x^2

Answer 18

@Michele_Laino

Answer 19

after the substitution above, you should get this:
\[\begin{gathered}
\int {\sqrt {1 + {t^2}} } \;dt = {\int {\left( {\cosh \theta } \right)} ^2}\;d\theta = \frac{{\theta + \sinh \theta \cosh \theta }}{2} = \hfill \\
= \left. {\frac{{\ln \left[ {\sqrt {1 + {t^2}} + t} \right] + t\sqrt {1 + {t^2}} }}{2}} \right|_1^x \hfill \\
\end{gathered} \]

Answer 20

Please continue, I think it is the right result, not the final result, of course.

Answer 21

1?

Answer 22

no, don't think, since the final answer must contain the x-variable

Answer 23

hint: if \[F(t) = \frac{{\ln \left[ {\sqrt {1 + {t^2}} + t} \right] + t\sqrt {1 + {t^2}} }}{2}\]
then you have evaluate this difference:\[F\left( x \right) - F\left( 1 \right)\]

Answer 24

this makes no sense, you said it was not , sqrt of 1+x^2??

Answer 25

those 2 are the only answers i get..

Answer 26

I said that \[\sqrt {1 + {x^2}} \] is not your answer

Answer 27

and you said it was not,

Answer 28

those are the only 2 answers i get :/

Answer 29

really your integration is not easy, so you have to be patient, please evaluate the difference above using my primitive function, and you will get your answer

Answer 30

i did!

Answer 31

what is it?

Answer 32

but if it is wrong, then i dont get why im getting this

Answer 33

please wait, I'm checking your expression

Answer 34

:)

Answer 35

I got this answer:
\[\ln \sqrt {\frac{{\sqrt {1 + {x^2}} + x}}{{1 + \sqrt 2 }}} - \left( {\frac{1}{{\sqrt 2 }} - \frac{{x\sqrt {1 + {x^2}} }}{2}} \right)\]

Answer 36

so it would be none of the above??

Answer 37

please wait, I try with another method

Answer 38

:)

Answer 39

Sorry, since your question requests the derivative of the function F(x), not the evaluation of the integral. So in order to get your answer, you have to apply the fundamental theorem of integral calculus

Answer 40

ohh thats why i was so confused on your work hahaha

Answer 41

That theorem states that the first derivative of F(x) is equal to the integrand function, namely:
\[F'\left( x \right) = \sqrt {1 + {x^2}} \]

Answer 42

So what is your option?

Answer 43

thats what i did and got the first one, A

Answer 44

are you sure?

Answer 45

please look at my function F ' (x)

Answer 46

oh wait! i meant my first answer, sorry i mixed it up ahhaha i put that in the very beginning (:

Answer 47

thanks(:

Answer 48

thanks! Please it is necessary that you study the fundamental theorem of integral calculus