The value of $\int\sin^{-1}\sqrt{(\frac{x}{a+x})}dx$ is:

$(a+x)\tan^{-1}\sqrt{(\frac{x}{a})}-\sqrt{ax}+C$

Put $x=a\tan^2θ$

$∴dx=2a\tan θ\sec^2θ\,dθ$  $∴\int\sin^{-1}\sqrt{(\frac{a\tan^2θ}{a\sec^2θ})}.2a\tan θ\sec^2θ\,dθ$

$=\int\sin^{-1}(\sin θ).2a\tan θ\sec^2θ\,dθ=a\int θ(2\tan θ\sec^2θ)dθ$

Integrate by parts

$I=a[θ\sec^2θ-\int\sec^2θ.1\,dθ]+C$  $[2\tan θ\sec^2θ\,dθ=\int 2\sec θ(\sec θ\tan θ\,dθ)=2\frac{\sec^2θ}{2}]$

$=a[θ(1+\tan^2θ)-\tan θ]=a[(1+\frac{x}{a})\tan^{-1}\sqrt{(\frac{x}{a})}-\sqrt{(\frac{x}{a})}]=(a+x)\tan^{-1}\sqrt{(\frac{x}{a})}-\sqrt{ax}+C$