<s>[INST] How to integrate $\int \frac{dx}{x\sqrt{x^4-1}}$? $$\int \dfrac{dx}{x\sqrt{x^4-1}}$$
I need to solve this integration.
I solved and got $\dfrac12\tan^{-1}(\sqrt{x^4-1}) + C$, however the answer given in my textbook is $\dfrac12\sec^{-1}(x^2) + C$
How can I prove that both quantities are equal? Is there something wrong with my answer?
EDIT:
Here's my work:
$$\int\dfrac{dx}{x\sqrt{x^4-1}}= \dfrac{1}{4}\int\dfrac{4x^3 dx}{x^4\sqrt{x^4-1}}$$
Let $x^4 - 1 = t^2$
$$\dfrac{1}{2}\int\dfrac{dx}{1 + t^2}$$
$$\dfrac12 \tan^{-1}(\sqrt{x^4 -1 }) + C$$
 [/INST] I tell my students that inverse trig functions are angles.  So if you write
$$\tan^{-1}\sqrt{x^4-1} = \theta,$$
then
$$\tan\theta = \sqrt{x^4-1}.$$
A right triangle that tells this story has $\theta$ as one angle, $\sqrt{x^4-1}$ as the opposite side and $1$ as the adjacent side.  Using Pythagorean theorem we can work out the length $c$ of the hypotenuse:
$$(\sqrt{x^4-1})^2+1^2 = c^2$$
which shows that $c=x^2$.
So $\sec \theta = x^2/1$, that is $\sec^{-1}(x^2) = \theta = \tan^{-1}( \sqrt{x^4-1}).$
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