Let $K>L>0$. I would like to find a good upper bound for the integral $$\int_0^L \sqrt{x \left(1 + \frac{1}{K-x}\right)} \,dx.$$ An explicit expression for the antiderivative would have to involve elliptic functions; thus, if we want to stick to simple expressions, a bound is the best we can do.

One obvious approach goes as follows: $$\begin{aligned} \int_0^L \sqrt{x \left(1 + \frac{1}{K-x}\right)} \, dx &\leq \int_0^L \sqrt{x} \left(1 + \frac{1/2}{K-x}\right)\, dx\\ &\leq \frac{2}{3} L^{3/2} + \frac{1}{2} \sqrt{\int_0^L x \,dx \cdot \int_0^L \frac{dx}{(K-x)^2}}\\ &= \frac{2}{3} L^{3/2} + \frac{1}{\sqrt{8}} \frac{L^{3/2}}{\sqrt{K (K-L)}} \end{aligned}$$ The second inequality (Cauchy-Schwarz) does not seem too bad, though one can easily improve on the constant $1/\sqrt{8}$ by proceeding more directly.

I dislike the first step, however, as it makes the integral diverge as $L\to K^-$, whereas the original integral did not.

What other simple approximations are there? Is it obvious that a bound of type $(2/3) L^{3/2} + O(L^{3/2}/K)$, say, could not be valid?