How to deal w/ inverse trig function within ∫(x)/((x^2-4)(√(x^4 -8)))?

[LearninDaMath]

Evaluate the integral \int \frac{x}{(x^2 - 4) \sqrt{x^4 - 8x^2}} \, dx

by making the substitution u = x^{2}

Homework Equations

The Attempt at a Solution

u = x^{2} - 4 so \frac{du}{2}= xdx

\int \frac{1}{(x^2 - 4) \sqrt{x^4 - 8x^2}} \, xdx\frac{1}{2} \int \frac{1}{(u) \sqrt{x^4 - 8x^2}} \, du\frac{1}{2} \int \frac{1}{(u)} \frac{1}{\sqrt{x^4 - 8x^2}} \, du

\frac{1}{2} \int \frac{1}{(u)} \frac{1}{|x|\sqrt{x^2 - 8}} \, duI haven't yet committed all of the inverse trig functions to memory but I have a table that shows dy/dsec^{-1}(x) = \frac{1}{|x| \sqrt{x^2 - 1}}

However, I do not know how to handle \frac{1}{|x| \sqrt{x^2 - 8}}

What does that equal and why? I mean what is the relationship between it and whatever it's antiderivative may be?

 

[vela]

Try doing it in two steps. Use the hint and make the first substitution u=x². Rewrite everything in terms of u. Then complete the square inside the radical and go from there.