Evaluate: ∫₂⁸ (√(10-x))/(√x + √(10-x)) dx

Let $I = \int_2^8 \frac{\sqrt{10-x}}{\sqrt{x}+\sqrt{10-x}}\,dx$. Using $\int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx$ with $a+b = 10$: $I = \int_2^8 \frac{\sqrt{x}}{\sqrt{10-x}+\sqrt{x}}\,dx$ Adding: $2I = \int_2^8 1\,dx = 6$ Hence $I = 3$.