Not Too Famous Integral

We can multiply and divide by the conjugate of the denominator to get;

$\begin{aligned}\int_0^3\sqrt{\dfrac{3-x}{3+x}}\, dx &=\int_0^3\sqrt{\dfrac{3-x}{3+x}}\times\sqrt{\dfrac{3-x}{3-x}}\, dx\\ &=\int_0^3\dfrac{3-x}{\sqrt{3^2-x^2}}\, dx\\ &=\int_0^3\dfrac{3}{\sqrt{3^2-x^2}}\, dx - \int_0^3\dfrac{x}{\sqrt{3^2-x^2}}\,dx\\ &=3\sin^{-1}\left(\dfrac x3\right)_0^3+\dfrac 12\int_{3^2}^0\dfrac{1}{\sqrt u}\, du\\ &=\dfrac{3\pi}{2}+\sqrt u|_{3^2}^{0}\\ &=3\left[\dfrac{\pi}{2}-1\right] \end{aligned}$

Thus, $A=3$ , $B=2$ and $C=1$ ; Therefore $A+B+C=\boxed 6$ .

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Note: This result can be generalised to any positive real number $a$ ;

$\int_0^a\sqrt{\dfrac{a-x}{a+x}}\, dx = a\left[\dfrac \pi 2-1\right]$