The value of the integral ∫limits_3^6 (√x)/(√(9-x)+√x) d x, is

$I=\int\limits_3^6 \frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}} d x$   ...(1)

so $I=\int\limits_3^6\frac{\sqrt{9-x}}{\sqrt{9-x}+\sqrt{x}}dx$   ...(2)

Eq. (1) + Eq. (2)

$⇒2I=\int\limits_3^61dx=3$

$I=\frac{3}{2}$