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 $x^4 - 1 = t^2$
$$\dfrac{1}{2}\int\dfrac{dx}{1 + t^2}$$
$$\dfrac12 \tan^{-1}(\sqrt{x^4 -1 }) + C$$
 A: 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 $\sec^{-1}(x^2) = \theta = \tan^{-1}( \sqrt{x^4-1}).$
A: The answer was already given by B. Goddard, here I try to give a visual answer:

A: I think your problem is already solved with B. Goddard's answer, Here's an easy way to solve the given integral.

We have,
$$\int\dfrac{dx}{x\sqrt{x^4-1}}$$
Let $x^2 = \sec(\theta) \implies 2x \, dx = \sec(\theta) \tan(\theta) \, d\theta$
Which further implies that, $dx = \dfrac{\sqrt{\sec(\theta)}\tan(\theta)}{2}\, d\theta$
After substitution, the given integral changes to:
$$\dfrac12\int\dfrac{\sqrt{\sec(\theta)}\tan(\theta)}{\sqrt{\sec(\theta)}\cdot\sqrt{\sec^2(\theta) - 1}}\, d\theta$$
$$=\dfrac12\int\dfrac{\sqrt{\sec(\theta)}\tan(\theta)}{\sqrt{\sec(\theta)}\cdot\sqrt{\tan^2(\theta)}}\, d\theta$$
$$=\dfrac12\int\dfrac{\sqrt{\sec(\theta)}\tan(\theta)}{\sqrt{\sec(\theta)}\tan(\theta)}\, d\theta$$
$$=\dfrac12\int d\theta$$
$$=\dfrac12\theta + C$$
$$=\boxed{\dfrac12\sec^{-1}(x^2) + C}$$
A: There is nothing wrong with your answer. The antiderivative you found is correct.
Your book's answer is also correct. The thing is that
$$\frac{1}{2}\tan^{-1}(\sqrt{x^4-1})=\frac{1}{2}\sec^{-1}(x^2)$$
If you don't believe me, see their graphs. The graphs of $\frac{1}{2}\tan^{-1}(\sqrt{x^4-1})$ and $\frac{1}{2}\sec^{-1}(x^2)$ are exactly the same.
Edit:
In order to prove that the quantities are indeed equal see @Zaragosa's answer.

PS: Whenever in doubt about calculus, use the derivative-calculator or the integral-calculator.
