TITLE: Integrating $ \int_2^4 \frac{ \sqrt{\ln(9-x)} }{ \sqrt{\ln(9-x)}+\sqrt{\ln(x+3)} } dx. $
QUESTION [8 upvotes]: Compute
$$
\int_2^4 \frac{ \sqrt{\ln(9-x)}      }{ \sqrt{\ln(9-x)}+\sqrt{\ln(x+3)}        } dx.
$$
I am not sure how to start this one...I am thinking of a substitution to get started.

REPLY [4 votes]: $\newcommand{\+}{^{\dagger}}
 \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\down}{\downarrow}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\isdiv}{\,\left.\right\vert\,}
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
 \newcommand{\wt}[1]{\widetilde{#1}}$
\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}