\( \int _{}^{}\frac{1}{x+\sqrt {x^{2}-1}} d x \)

To solve the integral \( \int \frac{1}{x+\sqrt{x^2-1}} \, dx \), we can use a clever substitution that simplifies our integrand. Notice that the expression \( x+\sqrt{x^2-1} \) can be related to the hyperbolic sine function. Let's make the substitution \( x = \cosh(t) \). With this substitution, we have \( dx = \sinh(t) \, dt \) and notice that \( \sqrt{x^2-1} = \sqrt{\cosh^2(t) - 1} = \sinh(t) \). Rewriting the integral, we get: \[ \int \frac{1}{\cosh(t) + \sinh(t)} \sinh(t) \, dt \] Now, simplifying the integrand gives: \[ \int \frac{\sinh(t)}{\cosh(t) + \sinh(t)} \, dt \] Using the identity \( \cosh(t) + \sinh(t) = e^t \), the integral simplifies to: \[ \int \frac{\sinh(t)}{e^t} \, dt = \int \frac{1}{2}(e^t - e^{-t}) e^{-t} \, dt = \frac{1}{2} \int (1 - e^{-2t}) \, dt \] Integrating gives: \[ \frac{1}{2} \left( t - \frac{1}{2} e^{-2t} \right) + C \] Now we just need to convert back to the original variable using \( t = \text{arcosh}(x) \): Thus, the solution can be rewritten in terms of \( x \) as: \[ \frac{1}{2} \left( \text{arcosh}(x) - \frac{1}{2(x^2 - 1)} \right) + C \] Now you have a comprehensive way to evaluate this integral!