MHB How Do You Solve This Complex Integral Problem?

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The integral problem presented for evaluation is ∫_{2}^{4} (√ln(9-x))/(√ln(9-x)+√ln(x+3)) dx. Participants are encouraged to refer to the Problem of the Week guidelines for submission procedures. Several members, including Opalg, lfdahl, kaliprasad, and greg1313, successfully provided correct solutions. The discussion highlights the collaborative effort in solving complex integral problems. Engaging with such challenges enhances mathematical skills and community interaction.
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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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Congratulations to the following members for their correct solution: (Cool)

1. Opalg
2. lfdahl
3. kaliprasad
4. greg1313Solution from Opalg:
$$\int_{2}^{4} \frac{\sqrt{\ln(9-x)}}{\sqrt{\ln(9-x)}+\sqrt{\ln(x+3)}}\,dx = \int_{2}^{3} \frac{\sqrt{\ln(9-x)}}{\sqrt{\ln(9-x)}+\sqrt{\ln(x+3)}}\,dx + \int_{3}^{4} \frac{\sqrt{\ln(9-x)}}{\sqrt{\ln(9-x)}+\sqrt{\ln(x+3)}}\,dx$$ 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 $$ \begin{aligned} \int_{2}^{4} \frac{\sqrt{\ln(9-x)}}{\sqrt{\ln(9-x)}+\sqrt{\ln(x+3)}}\,dx &= \int_{2}^{3} \frac{\sqrt{\ln(9-x)}}{\sqrt{\ln(9-x)}+\sqrt{\ln(x+3)}}\,dx + \int_{2}^{3} \frac{\sqrt{\ln(x+3)}}{\sqrt{\ln(x+3)}+\sqrt{\ln(9-x)}}\,dx \\ &= \int_{2}^{3} \frac{\sqrt{\ln(9-x)} + \sqrt{\ln(x+3)}}{\sqrt{\ln(9-x)}+\sqrt{\ln(x+3)}}\,dx \\ & = \int_{2}^{3}1\,dx \\ & = {\Large 1} \end{aligned}$$
 
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