Is the following integration "trick" valid to reduce my integrand to a constant? $$\int_2^4 \frac{1}{\sqrt{\frac{\ln(3+x)}{\ln(9-x)}} +1}dx = 1$$
I noticed when $x$ went from $2$ to $4$, $3+x$ went from $5$ to $7$, and $9-x$ went from $7$ to $5$.
I noticed that if we reverse the interval of $9-x$ we obtain $-(9-x)$ and whose interval goes from $5$ to $7$. In short, I concluded that reversing the interval of $9-x$ yielded the expression $3+x$.
Therefore I allowed the above integral to be $$\int_2^4 \frac{1}{\sqrt{\frac{\ln(3+x)}{\ln(9-x)}} +1} dx = \int_2^4 \frac{1}{\sqrt{\frac{\ln(3+x)}{\ln(3+x)}} +1} dx.$$
Effectively, the logs cancelled and I was left with $\int_2^4 {1\over 1+1}dx =1$.  
 A: Of course not, at least not in general: the argument is wrong, even though the function is set up very cleverly.
$3+x$ and $9-x$ may cover the same interval, but for each $x$, these two expressions are different. They only coincide once, otherwise you get a fraction with different top and bottom side. Even more... even if you wanted to use symmetry to argue left and right half of the integral cancels somehow, $\ln x$ is not a linear function - its shape on the bottom and upper part of the interval is completely different. If you want to prove or disprove symmetry in this particular case, the secret will lie more deeply in the properties of the given function, not just in the reversal of the interval.
You could try to use this kind of thinking to find a substitution which makes the integration domain and the terms under the square root symmetric.
In particular: if we try $u=x-3$, we get
$$\int_{-1}^1 \frac{1}{\sqrt{\frac{\ln (6+u)}{\ln (6-u)}}+1}du$$
which looks a bit better. The real question is, if we exchange $u\iff -u$, does the integrand resemble the original one?
Hypothesis:
$$\frac{1}{\sqrt{\frac{\ln (6+u)}{\ln (6-u)}}+1}=1-\frac{1}{\sqrt{\frac{\ln (6-u)}{\ln (6+u)}}+1}$$
if $-6<u<6$. Can you proceed on your own?
A: Counterexample:
$$\int_2^4\frac{3+x}{9-x}dx=\int_2^4\left(1-\frac{6}{9-x}\right)dx=\left.\left(x+6\log(9-x)\right)\right|_2^4=2+6\log\frac37\ne\int_2^4dx.$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\int_{2}^{4}{1 \over \root{\ln\pars{3 + x}/\ln\pars{9 - x}} + 1 }\,\dd x =
\int_{2}^{4}{\root{\ln\pars{9 - x}} \over \root{\ln\pars{3 + x}} +
\root{\ln\pars{9 - x}}}\,\dd x
\\[1cm] = &\ 
{1 \over 2}\left\{\int_{2}^{4}{\root{\ln\pars{9 - x}} \over \root{\ln\pars{3 + x}} +
\root{\ln\pars{9 - x}}}\,\dd x\right.
\\[5mm] &\
\left. + \int_{2}^{4}{\root{\ln\pars{9 - \bracks{4 + 2 - x}}} \over
\root{\ln\pars{3 + \bracks{4 + 2 - x}}} +
\root{\ln\pars{9 - \bracks{4 + 2 - x}}}}\,\dd x\right\} =
{1 \over 2}\int_{2}^{4}\dd x =
\bbox[#ffe,10px,border:1px dotted navy]{\ds{1}}
\end{align}
