posted by Terry .
What is the integral of 1/(x√(x^2-4)) dx?
So I know that I am going to have to factor out the 4 and use the arcsec trig identity but I've having trouble getting to a viable u-substitution.
I've gotten to factoring like this so far:
√(x^2-4) = √[2^2(x^2/2^2 - 1)] = 2√((x/2)^2 - 1))
I could do
u = x/2
du = 1/2 dx
But that still leaves the x in front of the radical.
How can I solve this?
I'm extremely rusty at these, but I think it's possible you might be able to use the substitution x=2cosh(u), where cosh is the hyperbolic cosine. cosh²(u)-sinh²(u)=1, so sqrt(x²-4) = sqrt(4cosh²(u)-4) = 2sinh(u). dx/du=2sinh(u), so dx=2sinh(u)du, which cancels out with the square root in the denominator, leaving 2cosh(u) in the denominator to perform the integral on. I can't remember whether that's integrable or not: you'd need to look into that. Is that any help?