<s>[INST] Calculate: $\int_{2}^{7}\frac{\sqrt{9-x}}{\sqrt{x}+\sqrt{9-x}}dx $ How to calculate following integration?
$$\int_{2}^{7}\frac{\sqrt{9-x}}{\sqrt{x}+\sqrt{9-x}}dx $$
 [/INST] Replace $x$ by $9-x$. We then get
$$I = \underbrace{\int_2^7 \dfrac{\sqrt{9-x}}{\sqrt{x} + \sqrt{9-x}} dx = \int_2^7 \dfrac{\sqrt{x}}{\sqrt{x} + \sqrt{9-x}} dx}_{x \to 9-x}$$
Adding both we get that
$$2I = \int_2^7 dx = 5 \implies I = \dfrac52$$

In general, if we have
$$I = \int_a^b \dfrac{f(a+b-x)}{f(x) + f(a+b-x)} dx$$ where $f(y) + f(a+b-y)$ doesn't vanish in $[a,b]$ then replacing $x$ by $a+b-x$, we get
$$I = \int_a^b \dfrac{f(a+b-x)}{f(x) + f(a+b-x)} dx = \int_a^b \dfrac{f(x)}{f(x) + f(a+b-x)} dx$$
Adding both gives us
$$2I = \int_a^b dx \implies I = \dfrac{b-a}2$$
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