Explanation
Consider the given integral.
$$ I=\int{\sqrt{\dfrac{a+x}{a-x}}dx} $$
$$ I=\int{\sqrt{\dfrac{a+x}{a-x}}\times \sqrt{\dfrac{a+x}{a+x}}dx} $$
$$ I=\int{\dfrac{a+x}{\sqrt{{{a}^{2}}-{{x}^{2}}}}dx} $$
$$ I=a\int{\dfrac{1}{\sqrt{{{a}^{2}}-{{x}^{2}}}}dx}-\dfrac{1}{2}\int{\dfrac{-2x}{\sqrt{{{a}^{2}}-{{x}^{2}}}}dx} $$
$$ I=a{{\sin }^{-1}}\left( \dfrac{x}{a} \right)-\dfrac{1}{2}\int{\left\{ \dfrac{d}{dx}\left( {{a}^{2}}-{{x}^{2}} \right){{\left( {{a}^{2}}-{{x}^{2}} \right)}^{-1/2}} \right\}dx} $$
$$ I=a{{\sin }^{-1}}\left( \dfrac{x}{a} \right)-\dfrac{1}{2}\dfrac{{{\left( {{a}^{2}}-{{x}^{2}} \right)}^{1/2}}}{\left( 1/2 \right)}+C $$
$$ I=a{{\sin }^{-1}}\left( \dfrac{x}{a} \right)-\sqrt{{{a}^{2}}-{{x}^{2}}}+C $$
Hence, this is the correct answer.
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