Solving Integrals 1

$\begin{aligned} I & = \int_3^6 \frac {\sqrt x}{\sqrt{9-x}+\sqrt x}\ dx & \small \color{#3D99F6} \text{Using identity }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx \\ & = \frac 12 \int_3^6 \left(\frac {\sqrt x}{\sqrt{9-x}+\sqrt x} + \frac {\sqrt {9-x}}{\sqrt x + \sqrt{9-x}}\right) dx \\ & = \frac 12 \int_3^6 dx = \frac x2 \ \bigg|_3^6 = \frac 32 \end{aligned}$

Therefore, $A+B = 3+2 = \boxed{5}$ .

Substitution $x = 9 - u$ and renaming $u \rightarrow x$ will show us that

$I = \int_3^6 \frac{\sqrt{x}}{\sqrt{9-x} + \sqrt{x}} dx = \int_3^6 \frac{\sqrt{9-x}}{\sqrt{9-x} + \sqrt{x}} dx$

Then, sum up the two representations for the same integral to get

$2I = \int_3^6 \left( \frac{\sqrt{x}}{\sqrt{9-x} + \sqrt{x}} + \frac{\sqrt{9-x}}{\sqrt{9-x} + \sqrt{x}} \right) dx = \int_3^6 \frac{\sqrt{9-x} + \sqrt{x}}{\sqrt{9-x} + \sqrt{x}} dx = \int_3^6 dx = 3.$

$\int \frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}} \, dx \Rightarrow \frac{1}{4} \left(2 x-2 \sqrt{-(x-9) x}+9 \log (9-2 x)+18 \tanh ^{-1}\left(\frac{\sqrt{x}}{\sqrt{9-x}}\right)\right)$

Evaluated at $x=3$ : $\frac{1}{4} \left(-6 \sqrt{2}+6+9 \log \left(6 \sqrt{2}+9\right)\right)$

Evaluated at $x=6$ : $-\frac{3}{\sqrt{2}}+3+\frac{9}{4} \log \left(6 \sqrt{2}+9\right)$

The difference is $\frac32$ .