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 $\frac{du}{2}= xdx$

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


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


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

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


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

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.