If ∫ 2x - 1 dx = α _ e | 2x + β + 2x (1 + cos(1)/(β)x) | +...

Given,

$∫ \sqrt{\sec 2x - 1}dx = \alpha{\log}_{e}\left|\cos 2x + \beta + \sqrt{\cos 2x\left(1 + cos\dfrac{1}{\beta}x\right)}\right| + C$

Now solving $L . H . S$ we get,

$∫ \sqrt{\sec 2x - 1}dx = ∫ \sqrt{\dfrac{1 - \cos 2x}{\cos 2x}}dx$

$⇒ ∫ \sqrt{\sec 2x - 1}dx = \sqrt{2} ∫ \dfrac{\sin x}{\sqrt{2\cos^{2}x - 1}}dx$

Now let $\cos x = t ⇒ - sinxdx = dt$

$⇒ ∫ \sqrt{\sec 2x - 1}dx = - \sqrt{2} ∫ \dfrac{dt}{\sqrt{2{t}^{2} - 1}}$

$⇒ ∫ \sqrt{\sec 2x - 1}dx = - ln | \sqrt{2}\cos x + \sqrt{\cos 2x} | + C$

$= - \dfrac{1}{2}\ln \left|2\cos^{2}x + \cos 2x + 2\sqrt{\cos 2x} \cdot \sqrt{2}\cos x\right| + C$

$= - \dfrac{1}{2}\ln \left|\cos 2x + \dfrac{1}{2} + \sqrt{\cos 2x} \cdot \sqrt{1 + \cos 2x}\right| + C$

Now on comparing with $∫ \sqrt{\sec 2x - 1}dx = \alpha{\log}_{e}\left|\cos 2x + \beta + \sqrt{\cos 2x\left(1 + cos\dfrac{1}{\beta}x\right)}\right| + C$

We get, $\beta = \dfrac{1}{2}, \alpha = - \dfrac{1}{2} ⇒ \beta - \alpha = 1$