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

To solve the integral \[ I = \int \frac{1}{\cos^2(x) \sqrt{1 + \tan(x)}} \, dx, \] we can start by making a substitution. Let \[ u = \tan(x). \] Then, the derivative of \( u \) with respect to \( x \) is \[ \frac{du}{dx} = \sec^2(x) = \frac{1}{\cos^2(x)}. \] This implies that \[ dx = \cos^2(x) \, du. \] Next, we need to express \( \sqrt{1 + \tan(x)} \) in terms of \( u \): \[ \sqrt{1 + \tan(x)} = \sqrt{1 + u}. \] Now, substituting \( u \) into the integral, we have: \[ I = \int \frac{1}{\cos^2(x) \sqrt{1 + u}} \cdot \cos^2(x) \, du = \int \frac{1}{\sqrt{1 + u}} \, du. \] The integral \[ \int \frac{1}{\sqrt{1 + u}} \, du \] can be solved using the standard integral formula: \[ \int \frac{1}{\sqrt{a + u}} \, du = 2\sqrt{a + u} + C. \] In our case, \( a = 1 \), so we have: \[ \int \frac{1}{\sqrt{1 + u}} \, du = 2\sqrt{1 + u} + C. \] Now, substituting back \( u = \tan(x) \): \[ I = 2\sqrt{1 + \tan(x)} + C. \] Thus, the final result for the integral is: \[ \int \frac{1}{\cos^2(x) \sqrt{1 + \tan(x)}} \, dx = 2\sqrt{1 + \tan(x)} + C. \]