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 $\int_2^4 {1\over 1+1}dx =1$.  

Counterexample:
$$\int_2^4\frac{3+x}{9-x}dx=\int_2^4\left(1-\frac{6}{9-x}\right)dx=\left.\left(x+6\log(9-x)\right)\right|_2^4=2+6\log\frac37\ne\int_2^4dx.$$
