Question

Fun fun question: Find the value of
\[\int_2^4 \frac{\sqrt{\ln(9-x)}}{\sqrt{\ln(9-x)}+\sqrt{\ln(x+3)}} dx\]

Answer 1

This one is a tough one!

Answer 2

I've seen it somewhere that I don't remember. I suppose you can rewrite the function f(x) as f(a+b-x) and it will be easier.

Answer 3

And then it'd be \[\int\limits \frac{f(x)}{f(x)+f(6-x)}dx\]

Answer 4

Fun fun question, fun fun answer :D
\[f(x) = \sqrt{\ln (9-x)}\]a= 2, b=4
\[f(a+b-x) =f(4+2-x) = f(6-x)= \sqrt{\ln (9-(6-x)} = \sqrt{x+3}\]
\[I = \int_a^b \dfrac{f(a+b-x)}{f(x) + f(a+b-x)} dx = \int_a^b \dfrac{f(x)}{f(x) + f(a+b-x)} dx\]
\[I=\int_2^4 \frac{\sqrt{\ln(9-x)}}{\sqrt{\ln(9-x)}+\sqrt{\ln(x+3)}} dx=\int_2^4 \frac{\sqrt{\ln(x+3)}}{\sqrt{\ln(9-x)}+\sqrt{\ln(x+3)}} dx\]\[2I=\int_2^4 \frac{\sqrt{\ln(9-x)}}{\sqrt{\ln(9-x)}+\sqrt{\ln(x+3)}} +\int_2^4 \frac{\sqrt{\ln(x+3)}}{\sqrt{\ln(9-x)}+\sqrt{\ln(x+3)}} dx\]\[2I=\int_2^4 \frac{\sqrt{\ln(9-x)}+\sqrt{\ln(x+3)}}{\sqrt{\ln(9-x)}+\sqrt{\ln(x+3)}} dx\]\[2I=\int_2^41dx\]\[2I=\int_2^4 \frac{\sqrt{\ln(9-x)}+\sqrt{\ln(x+3)}}{\sqrt{\ln(9-x)}+\sqrt{\ln(x+3)}} dx\]\[2I=2\]\[I=1\]
So,
\[\int_2^4 \frac{\sqrt{\ln(9-x)}}{\sqrt{\ln(9-x)}+\sqrt{\ln(x+3)}} dx=1\]

Answer 5

*Fourth line should be \[f(a+b−x)=f(4+2−x)=f(6−x)=\sqrt{ln(9−(6−x)}=\sqrt{\ln(x+3)}\]