# Thread: One more integral help

1. ## One more integral help

I need help with this:

1/(x((x^2-4)^1/2))

One divided by (x times the square root of x^2-4)

A buddy got it to a couple of ln's but I used trig substitution.

I let:
x=2sec(theta)
dx=2sec(theta)tan(theta) dtheta

So I'm integrating:

2sec(theta)tan(theta)/2sec(theta)*(4sec^2(theta)-4)

Which the 2sec(theta) cancels right away, the 4 under the radical comes out as a 2, and the (sec^2(theta)-1) turns into a tan^2(theta). Then I took the square root of the tan^2(theta) which then cancels with the above tan(theta).

So I'm integrating:

1/2

Which I get 1/2(theta), which I turn into (1/2)arcsec(x/2)

Is that right?

2. Yes that is correct.

3. So I have to do the limit from 2 to Infinity. So I broke it from 2 to 4 and from 4 to Infinity and let t=2 and t=infinity in those, since it's double improper. I need to do:

lim (as t approaches 2 from the right) of
(1/2arcsec(2)-1/2arcsec(t/2))

+

lim (as t approaches infinity) of
(1/2arcsec(t/2)-1/2arcsec(2)

Right?

4. You would need to take the limit as you approch any point of discontinity.

5. He pointed us towards a problem in the Calculus book that said we had to since at 2, it's division by zero and you can't evaluate infinity.

How would you do it if you don't have to break it up?

6. Substitution $u^2=x^2-4$ also works nice here.