How Do You Solve This Complex Integral Problem?

[anemone]

Here is this week's POTW:

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Evaluate $$\int_{2}^{4} \frac{\sqrt{\ln(9-x)}}{\sqrt{\ln(9-x)}+\sqrt{\ln(x+3)}}\,dx$$.

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Remember to read the https://mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to https://mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Congratulations to the following members for their correct solution: (Cool)

1. Opalg
2. lfdahl
3. kaliprasad
4. greg1313Solution from Opalg:

Spoiler

$$\int_{2}^{4} \frac{\sqrt{\ln(9-x)}}{\sqrt{\ln(9-x)}+\sqrt{\ln(x+3)}}\,dx = \int_{2}^{3} \frac{\sqrt{\ln(9-x)}}{\sqrt{\ln(9-x)}+\sqrt{\ln(x+3)}}\,dx + \int_{3}^{4} \frac{\sqrt{\ln(9-x)}}{\sqrt{\ln(9-x)}+\sqrt{\ln(x+3)}}\,dx$$ In that last integral, make the substitution $y = 6-x$ (so that $9-x = y+3$ and $x+3 = 9-y$): $$\int_{3}^{4} \frac{\sqrt{\ln(9-x)}}{\sqrt{\ln(9-x)}+\sqrt{\ln(x+3)}}\,dx = \int_{3}^{2} \frac{\sqrt{\ln(y+3)}}{\sqrt{\ln(y+3)}+\sqrt{\ln(9-y)}}(-dy) = \int_{2}^{3} \frac{\sqrt{\ln(y+3)}}{\sqrt{\ln(y+3)}+\sqrt{\ln(9-y)}}\,dy$$ Now replace the dummy variable $y$ by $x$ to get $$ \begin{aligned} \int_{2}^{4} \frac{\sqrt{\ln(9-x)}}{\sqrt{\ln(9-x)}+\sqrt{\ln(x+3)}}\,dx &= \int_{2}^{3} \frac{\sqrt{\ln(9-x)}}{\sqrt{\ln(9-x)}+\sqrt{\ln(x+3)}}\,dx + \int_{2}^{3} \frac{\sqrt{\ln(x+3)}}{\sqrt{\ln(x+3)}+\sqrt{\ln(9-x)}}\,dx \\ &= \int_{2}^{3} \frac{\sqrt{\ln(9-x)} + \sqrt{\ln(x+3)}}{\sqrt{\ln(9-x)}+\sqrt{\ln(x+3)}}\,dx \\ & = \int_{2}^{3}1\,dx \\ & = {\Large 1} \end{aligned}$$