\( \int \frac { 1 } { x \sqrt { x ^ { 2 } - 1 } } d x = [ ? ] ^ { - 1 } x + C \)

To solve the integral \( \int \frac{1}{x \sqrt{x^2 - 1}} \, dx \), we can utilize a substitution technique. Knowing that the expression under the square root involves \( x^2 - 1 \), we can consider the substitution \( x = \sec(\theta) \). This gives us \( dx = \sec(\theta) \tan(\theta) \, d\theta \) and transforms our integral into a form involving trigonometric functions. After performing the integration and making appropriate substitutions back to the original variable, we find the solution to match the specified form in the question. The integral can be expressed in terms of hyperbolic functions too. Specifically, the integral evaluates to \( \ln | x + \sqrt{x^2 - 1} | + C \). So we have: The final answer is \( \ln | x + \sqrt{x^2 - 1} | = [\text{arcosh}(x)]^{-1} + C \). Now, let's get to the fun part about problem-solving and some helpful tips! When tackling integrals that involve square roots of polynomials, it’s always great to look for trigonometric or hyperbolic substitutions—especially when dealing with expressions like \( \sqrt{x^2-1} \) that lend themselves well to the secant function. A common mistake to avoid here is to forget to handle the absolute value when taking logarithms. Always remember, the domain of your original function can dictate the presence of absolute values in the final answer! Happy integrating!