Let $I =\int\limits_{2}^{8} \frac{\sqrt{10-x}}{\sqrt{x}+\sqrt{10-x}} d x \dots$(i)
$\Rightarrow I =\int\limits_{2}^{8} \frac{\sqrt{10-(10-x)}}{\sqrt{10-x}+\sqrt{10-(10-x)}} d x$
$\left[\because \int\limits_{a}^{b} f(x) d x=\int\limits_{a}^{b} f(a+b-x) d x\right]$
$\Rightarrow I=\int\limits_{2}^{8} \frac{\sqrt{x}}{\sqrt{10-x}+\sqrt{x}} d x\dots$(ii)
On adding Eqs. (i) and (ii), we get
$ 2 I=\int\limits_{2}^{8} \frac{\sqrt{10-x}+\sqrt{x}}{\sqrt{10-x}+\sqrt{x}} d x $
$\Rightarrow 2 I=\int\limits_{2}^{8} 1 d x=[x]_{2}^{8} $
$\Rightarrow 2 I=(8-2)=6 $
$\therefore I=\frac{6}{2}=3$