Question

How is the following improper integral convergent? When I do partial fraction decomposition, I end up getting ln's, and if I sub in infinity into ln, I will most certainly get infinity, thus it would be divergent....

\[\int\limits_{2}^{\infty}\frac{dx}{x^2 + 2x - 3}\]

Answer 1

@dpaInc

Answer 2

Well, looking at this, the function is smaller than 1/x^2 for x>2, so it must converge.

Answer 3

what is the partial fraction decomposition you get for this?

Answer 4

Rogue is correct.

Answer 5

the integral is definitely convergent - if you show me what you did then maybe I can spot where you went wrong?

Answer 6

\[\int\limits_{2}^{\infty} \frac {dx}{x^2 + 2x - 3} = \frac {1}{4} \int\limits_{2}^{\infty} \frac {1}{x-1} - \frac {1}{x+3} dx = \frac {1}{4} \lim_{b \rightarrow \infty} \left[ \ln (x - 1) - \ln (x+3) \right]_{2}^{b}\]\[\frac {1}{4} \lim_{b \rightarrow \infty} \left[ \ln \frac {x - 1}{x+3} \right]_{2}^{b} = \frac {1}{4} \ln \lim_{b \rightarrow \infty} \frac {b-1}{b+3} - \frac {1}{4} \times \ln \frac {1}{4} = \frac {1}{4} ( \ln 1 - \ln \frac {1}{4}) = \frac{1}{4} ( 0 + \ln 4) = \frac {1}{4} \ln 4\]

Answer 7

\[= \frac {1}{4} (0 + \ln 4) = \frac {\ln 4}{4}\]

Answer 8

oh well - I guess Rogue has down all the work for you

Answer 9

lol, sorry :3

Answer 10

although I believe Rogue's answer is not correct

Answer 11

Yup, seems like I made a mistake.

Answer 12

Yeah, I think there should be a 5 in there somewhere..

Answer 13

Yeah, I see now. It should be (ln 5) /4

Answer 14

When substituting 2 into x + 3, it should be 5, not 4, lol :3

Answer 15

Ok I have my attempt out and ready: @asnaseer

Answer 16

One esc

Answer 17

http://i1084.photobucket.com/albums/j409/QRAWarrior/MATA36/MATA36-A4-Attempt1.png

Answer 18

I end up with those ln's at the end and those do go to infinity.

Answer 19

Think you could help me with my Calc. question, Rogue?

Answer 20

Sure, hold on.

Answer 21

Apply L'hopital's rule QRAWarrior.

Answer 22

How though?

Answer 23

Or you can just use end behavior to find the limit.

Answer 24

you can combine the two limits to get what Rogue did above

Answer 25

I am going to try to combine the two limits and see what happens. Don't post any answers! I will be right back

Answer 26

I think the key also is to write the integral as \[(1/4)\ln[(x-1)/(x+3)]\]

Answer 27

i.e. do this:\[\frac {1}{4} \lim_{b \rightarrow \infty} \left[ \ln (x - 1) - \ln (x+3) \right]_{2}^{b}\]

Answer 28

which in turn simplifies to:\[\frac {1}{4} \lim_{b \rightarrow \infty} \left[ \ln \frac {x - 1}{x+3} \right]_{2}^{b}\]

Answer 29

as Rogue has shown above.

Answer 30

My answer is ln(5)/4.

Answer 31

I took the advice of joining the logarithms together

Answer 32

perfecto!

Answer 33

When I simplified the argument in the logarithm using the limit, I ended up getting ln|1|, which is 0.

Answer 34

This was not even the first time I ran into the ln's and I was confused as to why it is NOT divergent

Answer 35

Finally I figured it out with your guys' help

Answer 36

@Rogue I wil give you a medal, if you give asnaseer a medal. I want to give you both, but I am not able to dhtat

Answer 37

I am not here for medals

Answer 38

only to help :)

Answer 39

@Rogue I have one more Calc problem that involves ln's. Would you mind helping me out?

Answer 40

I have to go somewhere in a few minutes, sorry :(
Asnaseer may be able to help :)