Question

show that the integral of x/(x^2-x-3) from 4 to infinity is either finite or unbounded..

Answer 1

sorry, that's x^2-2x-3

Answer 2

I believe we first need to figure out the integral and then check the limits right?

Answer 3

okay, i jumped to fast... that (by partial fractions) is unbounded.... The real question is show that the integral of (x/(x^2-2x-3))^2 is finite or unbounded. I think we need to compare the integral with either 1/x (divergent) or 1/x^2 (convergent). I'm just having a little trouble with the comparison. I also know that it IS bounded and finite.

Answer 4

I will start with some steps, you interrupt me when anything seems strange:

\[\Large \int\frac{x}{x^2-2x-3}dx=\int \frac{x}{x^2-2x+1-1-3}dx\\
\Large \int \frac{x}{(x-1)^2-4}dx \]
which might be as well solved with an appropriate substitution u=x-1
\[\Large u=x-1 \longrightarrow du=dx \]
from which:

\[\Large \int \frac{u+1}{u^2-4}du=\int \frac{u+1}{(u-2)(u+2)}du \]

Answer 5

Clearly looks unbounded to me, the fractions will lead up to partial fractions, which will again lead up to the natural logarithm.

x=4 is defined, is a bound
x= \(\infty\) is not defined.

Answer 6

I did get a comparison (x/x^2-2x-3)^2 > (1/x^2) and 1/x^2 converges. But the desired intergrand is greater than 1/x^2 and that doesn't help.

Answer 7

if (x/(x^2-2x-3))^2 were < 1/x^2, I could establish that it must converge. But it's > (1/x^2)...

Answer 8

Hmm yes, I didn't read that part of your question when I started already to work out the integral. If I carry out the last few steps I obtain:
\[\Large \frac{u+1}{(u-2)(u+2)}=\frac{A}{u-2}+\frac{B}{u+2} \]

Cover up method, or regular solving shows that A=3/4 and B =1/4

so our result would be:
\[\Large \left.\frac{1}{4}\left(3\ln(x-3)+\ln(x+1)\right)\right|^{\infty}_4 = \infty \]

Answer 9

but if you need to solve it by comparing I would recommend you to await further responses by somebody else, I don't feel too comfortable with doing that, seen it before but lack of experience.

Answer 10

Gabriel's horn..... the curve 1/x from x=1 to infinity is infinite, BUT the rotation of 1/x about the x-axis from 1 to infinity has finite volume..... I know that has something to do with it.

Answer 11

Comparing with more primitive forms of an integral is a good idea to approach such a problem.

In the regular integral test as I know it, I just compute the integral and evaluate it, if it's infinity, the integral will diverge, therefore the given term does diverge, not converge.

Answer 12

if the denominator was x^2+2x+3, then (x/(x^2 +2x+3))^2 would be < (1/x)^2..... and I could show convergence. But it's >(1/x)^2.......... :-(