Integrating $ \int_2^4 \frac{ \sqrt{\ln(9-x)} }{ \sqrt{\ln(9-x)}+\sqrt{\ln(x+3)} } dx. $

Split the integral up at $x=3$ to get

$$\int_2^3 dx \frac{\sqrt{\log{(9-x)}}}{\sqrt{\log{(9-x)}}+\sqrt{\log{(3+x)}}}+\int_3^4 dx \frac{\sqrt{\log{(9-x)}}}{\sqrt{\log{(9-x)}}+\sqrt{\log{(3+x)}}}$$

In the second integral, sub $x=6-y$. Then add the 2 integrals together. The answer is $1$.

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ \begin{align} &\color{#00f}{\large\int_{2}^{4}{% \root{\ln\pars{9 - x}} \over \root{\ln\pars{9 - x}} + \root{\ln\pars{x + 3}}}\,\dd x} \\[5mm] =&\ \int_{-1}^{1}{% \root{\ln\pars{6 - x}} \over \root{\ln\pars{6 - x}} + \root{\ln\pars{x + 6}}}\,\dd x \\[8mm]= &\ \left[\int_{0}^{1}{% \root{\ln\pars{6 - x}} \over \root{\ln\pars{6 - x}} + \root{\ln\pars{x + 6}}}\,\dd x\right. \\[2mm] & \left. + \int_{0}^{1}{% \root{\ln\pars{6 + x}} \over \root{\ln\pars{6 + x}} + \root{\ln\pars{-x + 6}}} \,\dd x\right] = \int_{0}^{1}\,\dd x = \color{#00f}{\Large 1} \end{align}