\( \left\{ \begin{array} { l } { \int \sqrt { 1 + x ^ { 2 } } d x } \\ { x = \cosh ( \epsilon ) } \end{array} \right. \)

To solve the integral \( \int \sqrt{1 + x^2} \, dx \) using the substitution \( x = \cosh(\epsilon) \), we first compute \( dx \) which gives us \( dx = \sinh(\epsilon) \, d\epsilon \). This substitution transforms the integral into: \[ \int \sqrt{1 + \cosh^2(\epsilon)} \sinh(\epsilon) \, d\epsilon \] Using the identity \( 1 + \cosh^2(\epsilon) = \sinh^2(\epsilon) + \cosh^2(\epsilon) = \cosh^2(\epsilon) + 1 \), we get \( \sqrt{1 + x^2} = \sqrt{\cosh^2(\epsilon)} = \cosh(\epsilon) \). Thus, our integral can be rewritten as: \[ \int \cosh(\epsilon) \sinh(\epsilon) \, d\epsilon \] The integral of \( \cosh(\epsilon) \sinh(\epsilon) \) can be solved easily, leading to \( \frac{1}{2} \sinh^2(\epsilon) + C \). Lastly, substituting back for \( x \) gives the original integral in terms of \( x \).