Evaluate definite integrals using the tables of integrals

- anonymous

Evaluate definite integrals using the tables of integrals

- Stacey Warren - Expert brainly.com

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- jamiebookeater

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- anonymous

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- Mimi_x3

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

- anonymous

oh ya thanks

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## More answers

- Mimi_x3

can you do it?

- anonymous

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

- anonymous

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

- anonymous

|dw:1326786254854:dw|

- anonymous

is that correct mimi?

- Diyadiya

Right ! :)

- anonymous

Hey thanks diya/lena

- anonymous

:D

- Diyadiya

*Diya :D

- KatrinaKaif

Are you even in HS yet? O.o

- Diyadiya

WHo me?

- anonymous

who me?

- anonymous

that pic is of lena

- KatrinaKaif

Lol No, the question is to rld

- anonymous

lol I am in university LOL

- KatrinaKaif

My apologizes!

- Mimi_x3

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

- anonymous

That is alright i like to be young

- anonymous

lol no I am in a stupid one

- KatrinaKaif

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

- anonymous

i am not taht old I am 20

- KatrinaKaif

Lol Never saw "20" coming :P

- Mimi_x3

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

- anonymous

lol i am learning it in calc2

- anonymous

well it is diffin india u guys learn alot more

- Mimi_x3

lols, im not from india xD

- anonymous

so where ru from?

- Mimi_x3

i thought calc 2 was harder :/

- KatrinaKaif

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

- Mimi_x3

Australia.

- anonymous

it is . this is the stupid stuff

- Mimi_x3

that is 11th grade stuff for me.

- anonymous

lol idk i never learnt this in hs

- KatrinaKaif

Ah I see Coolio

- Mimi_x3

wow, HS must be easy for you!

- anonymous

lol we learnt diff thing

- anonymous

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

- anonymous

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

- KatrinaKaif

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

- Mimi_x3

same, i hate trig and geometry.

- anonymous

but neway this university is in the us

- anonymous

ad this was one tiny sectionin calc

- KatrinaKaif

Interesting..

- anonymous

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

- Mimi_x3

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

- anonymous

lol i am from canada

- KatrinaKaif

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

- KatrinaKaif

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

- anonymous

lol zed was herer and then he left lol

- anonymous

i'm back :)

- anonymous

lol

- KatrinaKaif

Wb

- anonymous

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

- KatrinaKaif

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

- anonymous

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

- anonymous

ok so i wont hold that against u hehe

- anonymous

The chat is neway boring and stupid

- Diyadiya

Answer is not complete you have to find the definite integral

- anonymous

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

- anonymous

Hey wld anyone know?

- anonymous

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

- anonymous

ln(0) is undefined

- anonymous

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

- anonymous

ya lol i dont know how to write equations lol

- anonymous

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

- anonymous

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

- Mimi_x3

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

- anonymous

http://www.wolframalpha.com/input/?i=1%2F{\sqrt{x^{2}%2B6x%2B10}}++dx
i haven't learnt hyperbolic or whatever these sinh and cosh are

- anonymous

rld? are you here

- Mimi_x3

the answer doesn't have to be in sinh

- anonymous

ya lol this just ln natural logarithm

- anonymous

idk where i went wron

- anonymous

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

- anonymous

okay, integral without putting in limits is ln( sqrt( (x + 6)^2 - (Sqrt(26)^2 ) + x + 6) now you can put in the limits
x^2 + 6x + 10 = (x + 6)^2 - (Sqrt(26)^2

- anonymous

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

- anonymous

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

- Mimi_x3

\[\ln(-1+3+\sqrt{(-1+3)^{2}+1} - \ln(-3+3+\sqrt{(-3+3)^{2}+1} \]
\[\ln(2+\sqrt{(4+1} )- \ln(-3+3+\sqrt{0+1} )\]
\[\ln(2+\sqrt{5} )-\ln(0+\sqrt{1} ) = \ln(2+\sqrt{5} )-\ln(1)\]

- anonymous

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

- anonymous

oh thanks ishaan and mimi

- anonymous

but the formula i gave you is right

- anonymous

Ishaan that is alright Thank you for trying :D

- anonymous

U always come when i need u

- anonymous

Thanks mimi

- Mimi_x3

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

- anonymous

lol i will recheck the steps

- anonymous

who is this guest watching us?

- anonymous

hello guest!

- anonymous

idk been here the whole time

- anonymous

spying u ishaan

- anonymous

x^2 + 6x + 10 = x^2 + 6x + 9 -9 + 10 = (x+3)^2 - 1
\[\log{\sqrt{x^2+ 6x+10}}+ x+3\] final answer i can't get wrong twice

- anonymous

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

- anonymous

yeah, a spy

- Mimi_x3

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

- anonymous

k i will recheck it

- anonymous

There are two spies now

- anonymous

\[\log{\sqrt{1 - 6 + 10} -1 + 3 } -(\log {\sqrt{ (-3)^2 + -6*3 + 10} -3 +3})\]
ln(sqrt5 + 2) - ( ln(sqrt(9 -18 + 10))

- Mimi_x3

lols, people just love your question.

- anonymous

that's it, i can't be wrong

- anonymous

lol ishann thankss :D

- Mimi_x3

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

- anonymous

two spies :-O hello guest or if you don't get guest then people who don't have an account or don't wanna sign in

- Mimi_x3

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

- anonymous

or they wanna spy on us

- anonymous

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

- anonymous

is* not us :-D

- anonymous

seems legit :-D

- anonymous

brb lunch time
instant noodles yumm :-D

- anonymous

ya got that thanks

- anonymous

you're welcome

- anonymous

going to bed it is 4 am

- anonymous

bye mimi

- anonymous

sweet dreams

- anonymous

lol

- Diyadiya

LoL

- anonymous

Thanks diya for reminding me

- Diyadiya

NoProblem (:

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