Question

Find the value of the constant C for which the integral \[\int\limits_{0}^{\infty}(x/x^2+1)−(C/3x+1)dx\] converges. Evaluate the integral for this value of C.

Answer 1

ok after integrating
\[= (1/2)\ln(x^{2}+1) - (C/3)\ln(3x+1)\]
now im having trouble finding the limit as x goes to infinity

Answer 2

you know to circumvent that indeterminate form, we can combine the fractions

Answer 3

Like this:

ln(x^2 + 1) C ln(3x + 1)
----------- - ------------ =
2 3

Answer 4

Then :
3 ln(x^2 + 1) - 2C ln(3x + 1)
-----------------------------
6

Answer 5

which is really:

(x^2 + 1)^3
1/6 ln -----------
(3x + 1)^2C

Answer 6

ok
\[\frac{3\ln(x^{2}+1) - 2C*\ln(3x+1)}{6}\]
yes

Answer 7

right, apply the log properties to what you wrote and you should get what i got:

(x^2 + 1)^3
1/6 ln -----------
(3x + 1)^2C

Answer 8

yes got that, now find a C that allows this to converge as x->inf

Answer 9

now for that to converge, we have to have the same powers of x on the top and bottom

Answer 10

Notice that the top has a polynomial whose degree is 6 and the bottom has a degree of 2c

Answer 11

oh right, haha thanks
c = 3
converges to ln(1/3)

Answer 12

to be the same, we simply have to solve this: 2C = 6

Answer 13

which is c=3

Answer 14

well
1/6 * ln(1/3)

Answer 15

hmm, that was a good problem

Answer 16

thanks for the help

Answer 17

np

Answer 18

what do you think?

Answer 19

pretty good for 1:00 in the morning

Answer 20

yeah i agree good calc problem...11 pm here so not as impressive

Answer 21

lol

Answer 22

were r u at? not North America i gues

Answer 23

guess*

Answer 24

no i am..AZ pacific time

Answer 25

2 hour time difference approx.