$\int {\sqrt {\frac{{a + x}}{{a - x}}} } dx$

}

Let $I = \int {\sqrt {\frac{{a + x}}{{a - x}}} } dx$

Let's put $x = {\mathop{\rm acos}\nolimits} 2\theta$

$\Rightarrow$ $dx = - a \cdot \sin 2\theta \cdot 2 \cdot d\theta$

therefore,$I = - 2\int {\sqrt {\frac{{a + a\cos 2\theta }}{{a - a\cos 2\theta }}} } \cdot a\sin 2\theta d\theta$

$= - 2a\int {\sqrt {\frac{{1 + \cos 2\theta }}{{1 - \cos 2\theta }}} } \sin 2\theta d\theta = - 2a\int {\sqrt {\frac{{2{{\cos }^2}\theta }}{{2{{\sin }^2}\theta }}} } \sin 2\theta d\theta$

$= - 2a\int {\cot } \theta \cdot \sin 2\theta d\theta = - 2a\int {\frac{{\cos \theta }}{{\sin \theta }}} \cdot 2\sin \theta \cdot \cos \theta d\theta$

$= - 4a\int {{{\cos }^2}} \theta d\theta = - 2a\int {(1 + \cos 2\theta )} d\theta$

$= - 2a\left[ {\theta + \frac{{\sin 2\theta }}{2}} \right] + C$

$= - 2a\left[ {\frac{1}{2}{{\cos }^{ - 1}}\frac{x}{a} + \frac{1}{2}\sqrt {1 - \frac{{{x^2}}}{{{a^2}}}} } \right] + C$

$= - a\left[ {{{\cos }^{ - 1}}\left( {\frac{x}{a}} \right) + \sqrt {1 - \frac{{{x^2}}}{{{a^2}}}} } \right] + C$