<s>[INST] Evaluating: $I_1 = \int\sin^{-1} \left(\sqrt{\frac{x}{x+a}}\;\right) dx$ $$I_1 =\int \sin^{-1} \left(\sqrt{\frac{x}{x+a}}\;\right) dx= ?$$
I tried substitution: $\sin^{-1} \left(\sqrt{\frac{x}{x+a}}\;\right) = \Xi$, but then I'm not able to do anything after the resulting integral. 
Could someone help? There must be a simple way to solve this...
 [/INST] Hint. Assume $a>0,\,x+a>0$. Integrating by parts gives
$$\int \sin^{-1} \left(\sqrt{\frac{x}{x+a}}\;\right) dx=x\sin^{-1} \left(\sqrt{\frac{x}{x+a}}\;\right) -\int x\left(  \frac1{2x} \frac{\sqrt{ax}}{x+a}\right)dx \tag1 $$ and the last integral is easy to evaluate
$$\int   \frac{\sqrt{x}}{x+a}\:dx=2\int   \frac{u^2}{u^2+a}\:du\quad (\sqrt{x}=u). \tag2$$ Bringing $(1)$ and $(2)$ together leads to

$$\int \sin^{-1} \left(\sqrt{\frac{x}{x+a}}\;\right) dx=x\sin^{-1} \left(\sqrt{\frac{x}{x+a}}\;\right) -\sqrt{ax}+ a\sqrt{x}\arctan \left( \sqrt{\frac{x}{a}}\right)+C.$$ 

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