Completing the square should work.
$\int_{0}^{u} \frac{1}{\sqrt{1-x} \sqrt{x-t}}=\int_{0}^{u} \frac{1}{ \sqrt{-x^2+(t+1)x-1}})\, dx =$ $=\int_{0}^{u} \frac{1}{ \sqrt{-x^2+(t+1)x-\frac{t^2+2t+1}{4}+\frac{t^2+2t-3}{4}}}\, dx $
Now, if $t^2+2t-3 <0$ the function is not well defined. Otherwise, setting $\alpha=\sqrt{\frac{t^2+2t-3}{4}}$ you have
$\int_0^u \frac{1}{\sqrt{\alpha^2-(x-\frac{t+1}{2})^2}}$
which can be calculated with the substitution $x-\frac{t+1}{2}=\alpha \sin (v)$. Of course you need to put some extra restrictions on $t$, to make sure that your function is defined on $[0,u]$.