Evaluate, on intervals where the real-valued integrands are defined, \[ \int\frac{dx}{x^2\sqrt{x^2-a^2}}, \] together with the special cases \(a=2\), \(a=3\), and \[ \int\frac{dx}{x^2\sqrt{4x^2-1}}. \] Also evaluate \[ \int\frac{dx}{2x\sqrt{x^2-16}}. \]
Solution
For \(a\ne0\), \[ \frac{d}{dx}\left(\frac{\sqrt{x^2-a^2}}{a^2x}\right) =\frac1{x^2\sqrt{x^2-a^2}}, \] so \[ \boxed{\int\frac{dx}{x^2\sqrt{x^2-a^2}} =\frac{\sqrt{x^2-a^2}}{a^2x}+C.} \] Hence \[ \boxed{\int\frac{dx}{x^2\sqrt{x^2-4}}=\frac{\sqrt{x^2-4}}{4x}+C,} \] \[ \boxed{\int\frac{dx}{x^2\sqrt{x^2-9}}=\frac{\sqrt{x^2-9}}{9x}+C.} \] Likewise, \[ \boxed{\int\frac{dx}{x^2\sqrt{4x^2-1}} =\frac{\sqrt{4x^2-1}}{x}+C,} \] as direct differentiation verifies.
For the final integral, put \(u=\sqrt{x^2-16}\). Since \(u\,du=x\,dx\) and \(x^2=u^2+16\), \[ \int\frac{dx}{2x\sqrt{x^2-16}} =\frac12\int\frac{du}{u^2+16} =\boxed{\frac18\arctan\!\left(\frac{\sqrt{x^2-16}}4\right)+C.} \] This form avoids the branch conventions inherent in \(\sec^{-1}(x/4)\).