A Nice Integral (1)

Use $\small{\displaystyle\int_a^bf(x)\mathrm{d}x=\displaystyle\int_a^bf(a+b-x)\mathrm{d}x}$

$\mathfrak W=\displaystyle\int_{2}^{4}\dfrac{\sqrt{ln(9-x)}}{\sqrt{ln(9-x)}+\sqrt{ln(x+3)}}$ $=\displaystyle\int_{2}^{4}\dfrac{\sqrt{ln(x+3)}}{\sqrt{ln(x+3)}+ \sqrt{ln(9-x)}}$

Adding the above two, we get :-

$2\mathfrak W=\displaystyle\int_2^4\mathrm{d}x$

$\Large \implies \mathfrak W=\boxed{\color{#3D99F6}{1}}~~~~~~~~~~~$

The integral is continuous at $[2,4]$ . Let $I$ be the value of the integral.

( $9-x$ ) and ( $x-3$ ) go from $7$ to $5$ and from $5$ to $7$ as $x \ goes\ from \ 2 \ to\ 4$ respectively;

This symmetry allows us to use the substitution $x=6-y$ ;

$I= \large \int_{2}^{4} \frac{\sqrt{ln(y+3)}}{\sqrt{ln(y+3)} +\sqrt{ln(9-y)}}$ $dx$ ; $\implies$

$2I= \large \int_{2}^{4} \frac{\sqrt{ln(x+3)}+\sqrt {ln(9-x)}}{\sqrt{ln(x+3)} +\sqrt{ln(9-x)}} dx$ = $\large \int_{2}^{4}dx$ = $4-2=2$ $\implies$

$\boxed{I=1}$ .