<s>[INST] 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.
 [/INST] Hint: 
Notice that $x^4-1 = (x^2-1)(x^2+1)$, and then split into 2 fractions:
$$\frac{\sqrt{x^2+1}-\sqrt{x^2-1}}{\sqrt{x^4-1}} = \frac{\sqrt{x^2+1}}{\sqrt{(x^2+1)(x^2-1)}} - \frac{\sqrt{x^2-1}}{\sqrt{(x^2+1)(x^2-1)}}$$
Then simplify and you get 2 easier integrals.
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