Question

Calc question!
How do i evaluate the integral of [2dx/(x^2+1)] over -infinity to -2?

Answer 1

over infinity you say?

Answer 2

oh you mean\[\int_{-\infty}^{-2}{2dx\over x^2+1}\]

Answer 3

yeah exactly! i totally forgot how to do improper integrals.

Answer 4

thanks for replying btw i was freakin out for a bit lol

Answer 5

we rewrite this as a limit-integral pair

Answer 6

oops its actually
x^2 - 1

Answer 7

\[\lim_{n\to-\infty}\int_{n}^{-2}{2dx\over x^2-1}\]I suppose we can do partial fractions to do this integral...

Answer 8

do you have any problems with this as an indefinite integral ?

Answer 9

its part of that section in the book

Answer 10

what section?
I'm asking if you can do the indefinite integral\[\int{2dx\over x^2-1}\]?

Answer 11

a little bit i want to say you use trig sub...

Answer 12

i need help with the entire problem im sorry haha

Answer 13

I would say partial fractions is the way to go

Answer 14

brb

Answer 15

ok

Answer 16

sorry about that, I hate long phone calls

Answer 17

anyway we would use partial fractions on this integral, are you familiar with that technique?

Answer 18

u -sub will work for this one

Answer 19

really?

Answer 20

\(u=x^2-1, du=2xdx\) etc

Answer 21

off hand i would say it diverges

Answer 22

it's 2dx on the top though

Answer 23

\[\int\frac{2xdx}{x^2-1}dx=\int\frac{du}{u}=\ln(u)\]

Answer 24

you are imagining an x :P

Answer 25

ooooooooooooh silly me you are right!!!

Answer 26

but yeah this integral still gets funky

Answer 27

the limit I'm not sure about...

Answer 28

no i think you may be able to do it.
\[\int \frac{dx}{x-1}-\int \frac{dx}{x+1}\]
\[\ln(x-1)-\ln(x+1)\]
\[\ln(\frac{x-1}{x+1})\]

Answer 29

I can do the integral just fine, but I'm unsure of the limit

Answer 30

now i think it is ok, because as x goes to minus infinity you get \(\ln(1)=0\) i gotta run

Answer 31

i don't think it's going to minus infinity, i think its going from minus infinity to minus 2

Answer 32

i might be wrong

Answer 33

it is... and I think satellite was in a hurry and messed up a little

Answer 34

first the integral:
partial fractions on the integrand\[{2\over x^2-1}={2\over(x+1)(x-1)}=\frac A{x+1}+\frac B{x-1}\]

Answer 35

@sk84 do you know this technique^^
?

Answer 36

yes

Answer 37

I think this diverges...
integral is easy, limit... not so much

Answer 38

so then you go ahead and get A and B while I think about the limit :)

Answer 39

ok lol

Answer 40

ok I think satellite is right about the limit

Answer 41

let me know when you get A and B

Answer 42

i think we have to do the lim of n->-2 from the left of ln (x-1) - ln (x+1)

Answer 43

i dunno though

Answer 44

you should write it as \[\ln\left(\frac{x-1}{x+1}\right)\] when you are evaluating the limits of integration

Answer 45

just plug in \(x=-2\)

Answer 46

...because as satellite said as \(x\to-\infty\) the expression becomes\[\ln1=0\]and we only have to evaluate at the other point
...right?

Answer 47

*\(n\to-\infty\)

Answer 48

\[\lim_{a\to-\infty}\int_{a}^{-2}{2dx\over x^2-1}\]

\[=\lim_{a\to-\infty}\left.\ln\left(\frac{x-1}{x+1}\right)\right|_a^{-2}\]

\[=\ln\left(\frac{-2-1}{-2+1}\right)-\lim_{a\to-\infty}\ln\left(\frac{a-1}{a+1}\right)\]

Answer 49

Just so you know Zarkon, I had figured it out already. I was just elsewhere for a moment...

Answer 50

good ;)

Answer 51

\[=\ln(3)\]

Answer 52

thanks guys! esp turing test and zarkon

Answer 53

welcome!