TITLE: Is the following integration "trick" valid to reduce my integrand to a constant?
QUESTION [29 upvotes]: $$\int_2^4 \frac{1}{\sqrt{\frac{\ln(3+x)}{\ln(9-x)}} +1}dx = 1$$
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 $\int_2^4 {1\over 1+1}dx =1$.

REPLY [2 votes]: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\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}