Yes that is correct.
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?
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?