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|dw:1326785742825:dw|

\[\int\limits\frac{1}{\sqrt{x^{2}+6x+10}} dx\]
Then complete the square.

oh ya thanks

can you do it?

well i will try lol
if i cant then i will page u

ok i got the answer u tell me if it is correct

|dw:1326786254854:dw|

is that correct mimi?

Right ! :)

Hey thanks diya/lena

:D

*Diya :D

Are you even in HS yet? O.o

WHo me?

who me?

that pic is of lena

Lol No, the question is to rld

lol I am in university LOL

My apologizes!

What..? university? is this uni lvl ? :/

That is alright i like to be young

lol no I am in a stupid one

Well then, we shall not be nostalgic about the past with our young minds ;)

i am not taht old I am 20

Lol Never saw "20" coming :P

lols, dont you learn that in HS? I learnt that last year :/

lol i am learning it in calc2

well it is diffin india u guys learn alot more

lols, im not from india xD

so where ru from?

i thought calc 2 was harder :/

I am in the US...Its..a 10th grade course..

Australia.

it is . this is the stupid stuff

that is 11th grade stuff for me.

lol idk i never learnt this in hs

Ah I see Coolio

wow, HS must be easy for you!

lol we learnt diff thing

we have one math course a year and that includes e/t.
algebra trig geometry and all th egood stuff

like we odnt have algebra 1 and 2 and all that stuff

Hmm, trig and Geo come in one pack for us.

same, i hate trig and geometry.

but neway this university is in the us

ad this was one tiny sectionin calc

Interesting..

katrina so y r u up now? it is so late

very interesting, maths in the US, sounds so easy xD

lol i am from canada

Depends Mimi. You have the choice to take the courses. So it varies from people.

It is a bit late.. Last minute procrastinator :P

lol zed was herer and then he left lol

i'm back :)

lol

Wb

I guess it is never too late i paged u on the caht but i guess u dont read the chat

Only if there was an "off" button to chat...

i usually do, but the internet is a bit slow right now

ok so i wont hold that against u hehe

The chat is neway boring and stupid

Answer is not complete you have to find the definite integral

lol i just realized that :D U r right on target

Hey wld anyone know?

If u can solve (ln(0)+1)?

ln(0) is undefined

Okay, \[\int_{-3}^{-1}\frac{1}{\sqrt{x^{2}+6x+10}} dx\]

ya lol i dont know how to write equations lol

but that wasn't my question ishaan. i want to know abt ln(0)

ln 0 isn't defined, i think you probably made some mistake

\[\ln(x+3+\sqrt{(x+3)^{2}+1} \]
then sub -1 then -3 in

rld? are you here

the answer doesn't have to be in sinh

ya lol this just ln natural logarithm

idk where i went wron

ln(-3+3)=ln(o)

it gets reduced to 1 upon sqrt{x^2 - a^2}

the integral of 1 upon sqrt(x^2-a^2) = log(sqrt(x^2-a^2) +x)

oh wait i did wrong mimi did right i got confused in completing squares

oh thanks ishaan and mimi

but the formula i gave you is right

Ishaan that is alright Thank you for trying :D

U always come when i need u

Thanks mimi

np, but i think that i did it wrong somewhere :/

lol i will recheck the steps

who is this guest watching us?

hello guest!

idk been here the whole time

spying u ishaan

\[\log({\sqrt{x^2+ 6x+10}}+ x+3)\]

yeah, a spy

the final is answer is right:
\[\ln(x+3+\sqrt{(x+3)^{2}+1} \]
but then the end is not i think,

k i will recheck it

There are two spies now

lols, people just love your question.

that's it, i can't be wrong

lol ishann thankss :D

i think that i got the same answer, scroll up!

they are just too lazy to sign up for an account :P

or they wanna spy on us

ln(sqrt(5) + 2) us the final answer
the other one gets to zero

is* not us :-D

seems legit :-D

brb lunch time
instant noodles yumm :-D

ya got that thanks

you're welcome

going to bed it is 4 am

bye mimi

sweet dreams

lol

LoL

Thanks diya for reminding me

NoProblem (: