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

by making the substitution [itex]u = x^{2}[/itex]

[itex]u = x^{2} - 4[/itex] so [itex]\frac{du}{2}= xdx[/itex]

[tex] \int \frac{1}{(x^2 - 4) \sqrt{x^4 - 8x^2}} \, xdx [/tex]

[tex]\frac{1}{2} \int \frac{1}{(u) \sqrt{x^4 - 8x^2}} \, du[/tex]

[tex]\frac{1}{2} \int \frac{1}{(u)} \frac{1}{\sqrt{x^4 - 8x^2}} \, du[/tex]

[tex]\frac{1}{2} \int \frac{1}{(u)} \frac{1}{|x|\sqrt{x^2 - 8}} \, du[/tex]

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

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

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

by making the substitution [itex]u = x^{2}[/itex]

## Homework Equations

## The Attempt at a Solution

[itex]u = x^{2} - 4[/itex] so [itex]\frac{du}{2}= xdx[/itex]

[tex] \int \frac{1}{(x^2 - 4) \sqrt{x^4 - 8x^2}} \, xdx [/tex]

[tex]\frac{1}{2} \int \frac{1}{(u) \sqrt{x^4 - 8x^2}} \, du[/tex]

[tex]\frac{1}{2} \int \frac{1}{(u)} \frac{1}{\sqrt{x^4 - 8x^2}} \, du[/tex]

[tex]\frac{1}{2} \int \frac{1}{(u)} \frac{1}{|x|\sqrt{x^2 - 8}} \, du[/tex]

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

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

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

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