Evaluation of $\int_{0}^{3a} \frac{x\sqrt{3a-x}}{\sqrt{a+x}} dx$ How can we evaluate following integral:
$$I=\int_{0}^{3a} \frac{x\sqrt{3a-x}}{\sqrt{a+x}} dx$$
I tried using the property $\int_{a}^{b} f(x) dx=\int_{a}^{b} f(a+b-x) dx$ but it not helping because of the denominator. How should I proceed?

Assume $a>0$. One may write
$$
\begin{align}
\int_{0}^{3a} \frac{x\sqrt{3a-x}}{\sqrt{a+x}} dx &=a^2\int_{0}^{3} \frac{t\sqrt{3-t}}{\sqrt{1+t}} dt \qquad (x=at)
\\\\&=2a^2\int_{1}^{2} (u^2-1)\sqrt{4-u^2}\: du \qquad \left(u=\sqrt{1+t}\:\right)
\\\\&=8a^2\int_{\pi/6}^{\pi/2} (3-4\cos^2 v)\cos^2 v\: dv \qquad \left(u=2\sin v\:\right)
\\\\&=\frac{3\sqrt{3}}2 a^2,
\end{align}
$$ where we have used 
$$
\cos^2 v=\frac{1+\cos(2v)}2
$$ twice.
