1. **State the problem:** We need to solve the integral $$\int \frac{1}{x \sqrt{x^4 - 4}} \, dx.$$\n\n2. **Identify substitution and formula:** Notice the expression under the square root is $x^4 - 4 = (x^2)^2 - 2^2$, which suggests a substitution involving $x^2$. Also, the integral resembles the form $$\int \frac{1}{x \sqrt{x^4 - a^4}} \, dx,$$ which can be solved using the substitution $u = x^2$.\n\n3. **Substitution:** Let $$u = x^2 \implies du = 2x \, dx \implies dx = \frac{du}{2x}.$$\n\n4. **Rewrite the integral in terms of $u$:**\n$$\int \frac{1}{x \sqrt{x^4 - 4}} \, dx = \int \frac{1}{x \sqrt{u^2 - 4}} \cdot \frac{du}{2x} = \int \frac{1}{2x^2 \sqrt{u^2 - 4}} \, du.$$\nSince $x^2 = u$, this becomes\n$$\int \frac{1}{2u \sqrt{u^2 - 4}} \, du.$$\n\n5. **Simplify the integral:**\n$$\int \frac{1}{2u \sqrt{u^2 - 4}} \, du = \frac{1}{2} \int \frac{1}{u \sqrt{u^2 - 4}} \, du.$$\n\n6. **Recognize the integral form:** The integral $$\int \frac{1}{u \sqrt{u^2 - a^2}} \, du$$ has the solution $$\frac{1}{a} \operatorname{arcsec} \left| \frac{u}{a} \right| + C.$$\nHere, $a = 2$.\n\n7. **Apply the formula:**\n$$\frac{1}{2} \int \frac{1}{u \sqrt{u^2 - 4}} \, du = \frac{1}{2} \cdot \frac{1}{2} \operatorname{arcsec} \left| \frac{u}{2} \right| + C = \frac{1}{4} \operatorname{arcsec} \left| \frac{u}{2} \right| + C.$$\n\n8. **Back-substitute $u = x^2$:**\n$$\int \frac{1}{x \sqrt{x^4 - 4}} \, dx = \frac{1}{4} \operatorname{arcsec} \left| \frac{x^2}{2} \right| + C.$$\n\n**Final answer:** $$\boxed{\int \frac{1}{x \sqrt{x^4 - 4}} \, dx = \frac{1}{4} \operatorname{arcsec} \left| \frac{x^2}{2} \right| + C}.$$