Why does WolframAlpha's expression for $\int\frac{dx}{x\sqrt{x^4-4}}$ disagree with my own? $$\int\frac{1}{x\sqrt{x^4-4}}$$
My teacher gave us these notes and I'm unsure if they're correct. Wolfram gives a different answer, and when I derive I might have messed up.
Thanks.

 A: Differentiating both the expression $$\frac{1}{4} \text{arcsec } \frac{x^2}{2} + C$$ derived in the question and the expression $$-\frac{1}{4} \arctan \frac{2}{\sqrt{x^4 - 4}} + C$$ given by WolframAlpha yields the original integrand, so both are correct. We can use a little easy trigonometric to see how this can be:
At least when $u \geq a > 0$, the quantity $$\text{arcsec } \frac{u}{a}$$ is (more or less by definition) the measure of the red angle denoted $\color{red}{\theta}$ in the below diagram.

On the other hand, we could just as well (again by definition) write $\color{red}{\theta}$ as $$\arctan \frac{\sqrt{u^2 - a^2}}{a},$$ and using the identity $$\arctan v + \arctan \frac{1}{v} = \frac{\pi}{2}$$ (which holds for $v > 0$, as similar considerations involving an appropriately labeled right triangle reveal) gives that we can also write $\color{red}{\theta} = \text{arcsec } \frac{u}{a}$ as $$\frac{\pi}{2} - \arctan \frac{a}{\sqrt{u^2 - a^2}}.$$
Now, taking $u = x^2$ and $a = 2$ as in the derivative in the question gives
$$\int \frac{dx}{x \sqrt{x^4 - 4}} = \text{arcsec } \frac{x^2}{2} + C = \frac{1}{4}\left(\frac{\pi}{2} - \arctan \frac{2}{\sqrt{x^4 - 4}}\right) + C.$$
Distributing the factor of $\frac{1}{4}$ and absorbing the constant $\frac{\pi}{8}$ into the constant of integration shows that the derived answer really does concide with WolframAlpha's answer.
One can see we could just as well have written $\color{red}{\theta}$ in terms of any of the other four inverse trigonometric functions, and each of these would have lead to an equally valid general expression for an antiderivative. These considerations also apply to a wide class of integrals that we solve using trigonometric substitution. (One should take the usual care required when working with inverse trigonometric functions, especially when working outside the "principal branch", as this can affect the antiderivative expressions in nonobvious ways.)
A: It's easy to check the correctness of an antiderivative by differentiating:
$$ \dfrac{d}{dx} \dfrac{1}{4} \text{arcsec}(x^2/2) = \dfrac{1}{x^3 \sqrt{1 - 4/x^4}} = \dfrac{1}{x \sqrt{x^4 - 4}}$$
So $\dfrac{1}{4} \text{arcsec}(x^2/2)+C$ is indeed an antiderivative of 
$1/(x \sqrt{x^4-4})$.  
I might note for completeness, however, that the domain of both 
the function and its antiderivative (assuming you're not interested in complex values) is disconnected: $(-\infty, -\sqrt{2}) \cup (\sqrt{2},\infty)$, and thus the "constant" $C$ might be different in the two intervals.
A: Let $$\displaystyle I = \int \frac{1}{x\sqrt{x^4-4}}dx\;,$$ Let $x^2=2\sec \phi\;,$ Then $xdx = \sec \phi\cdot \tan \phi d\phi$
So Integral $$\displaystyle I = \int\frac{x}{x^2\sqrt{x^4-4}}dx = \int\frac{\sec\phi\cdot \tan \phi}{2\sec \phi\sqrt{4(\sec^2 \phi-1)}}d\phi$$
Using the formula $\bullet\; \sec^2 \phi = 1+\tan^2 \phi\Rightarrow \sec^2 \phi = 1+\tan^2 \phi$
So $$\displaystyle I = \frac{1}{4}\int\frac{\sec \phi\cdot \tan \phi}{\sec \phi\cdot \tan \phi}d\phi = \frac{1}{4}\int 1d\phi = \frac{1}{4}\phi+\mathcal{C}$$
so we get $$\displaystyle I = \frac{1}{4}\sec^{-1}\left(\frac{x^2}{2}\right)+\mathcal{C}$$
so Your answer is Right.
