# Integral - how would you compute this expression?

$$\int\frac{\sqrt{x^2+1}-\sqrt{x^2-1}}{\sqrt{x^4-1}}dx$$

I tried many types of substitutions, but I got nowhere.

-

Hint: Notice that $x^4-1 = (x^2-1)(x^2+1)$, and then split into 2 fractions:
$$\frac{\sqrt{x^2+1}-\sqrt{x^2-1}}{\sqrt{x^4-1}} = \frac{\sqrt{x^2+1}}{\sqrt{(x^2+1)(x^2-1)}} - \frac{\sqrt{x^2-1}}{\sqrt{(x^2+1)(x^2-1)}}$$
HINT: Make use of $x^4 - 1 = (x^2+1)(x^2-1)$ to reduce the integral at hand to two simpler integrals.