How to calculate $\int_2^4 \frac{\sqrt{\ln(9-x)}}{\sqrt{\ln(9-x)} + \sqrt{\ln(3+x)}} \space dx$

Question

How to calculate $$\int_2^4 \frac{\sqrt{\ln(9-x)}}{\sqrt{\ln(9-x)} + \sqrt{\ln(3+x)}} \space\mathrm{d}x$$

Answer 1

Short answer: $I=1$.

Proof: Note that \begin{align*} I := \int_2^4 \frac{\sqrt{\ln(9-x)}}{\sqrt{\ln(9-x)} + \sqrt{\ln(3+x)}} \space dx\\ \overset{\text{substitute } v := 3-x}{=} - \int_{-1}^1 -\frac{\sqrt{\ln(6+v)}} {\sqrt{\ln(6+v)} + \sqrt{\ln(6-v)}} \space dx\\ \overset{\text{substitute } u := -v}{=} \int_{-1}^1 -\frac{\sqrt{\ln(6-u)}}{\sqrt{\ln(6+u)} + \sqrt{\ln(6-u)}} \space dx \end{align*}

Thus, $2\cdot I = \displaystyle\int_{-1}^1 \frac{\sqrt{\ln(6-x)}+\sqrt{\ln(6+x)}}{\sqrt{\ln(6+x)} + \sqrt{\ln(6-x)}} \space dx = \int_{-1}^1 1 = 2$, i.e. $I = 1$

Answer 2

$$\mathbf I =\int_2^4 \frac{\sqrt{\ln(9-x)}}{\sqrt{\ln(9-x)} + \sqrt{\ln(3+x)}}\mathrm{dx}=\int_2^4 \frac{\sqrt{\ln(9-(4+2-x))}}{\sqrt{\ln(9-(4+2-x))} + \sqrt{\ln(3+(4+2-x))}}\mathrm{dx} \qquad\text{(Why?)}$$ So $\mathbf I =\displaystyle\int_2^4 \frac{\sqrt{\ln(3+x)}}{\sqrt{\ln(3+x)} + \sqrt{\ln(9-x)}}\mathrm{dx}$.

Then $2\mathbf I =\int_2^4 \mathrm{dx}=2$.