Integrating $ \int_2^4 \frac{ \sqrt{\ln(9-x)} }{ \sqrt{\ln(9-x)}+\sqrt{\ln(x+3)} } dx. $ 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.  
 A: 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 $1$.
A: $\newcommand{\+}{^{\dagger}}
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\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}
