Finding value of $\displaystyle \int \sqrt{x+\sqrt{kx-x^2}}dx$
My Try: Let $\displaystyle I = \sqrt{x+\sqrt{kx-x^2}}$. Substitute $kx-x^2=t^2$
Then $(k-2x)dx = dt$
$\Rightarrow \displaystyle I = \int \sqrt{x+t}\; \times \frac{1}{k-2x}dt$
Now $\displaystyle x^2-kx+t^2=0\Rightarrow x=\frac{k\pm \sqrt{k^2-4t^2}}{2}$
$\displaystyle \Rightarrow I = \pm \int \sqrt{\frac{k\pm \sqrt{k^2-4t^2}}{2}+t}\times \frac{1}{\sqrt{k^2-4t^2}}dt$
Could some help me how to solve it
please help me , Thanks