1. ## Challenging Integral

I was tutoring someone in Calculus today, and they had an extra credit assignment to do. These integrals appeared on past PUTNAM exams, and they can get help on these. However, I've been trying to work this one out and I (and all the other tutors I work with) couldn't figure it out.

So for the sake of me giving a hint or two to this person, I would like some guidance on how to tackle this beast:

$\int_2^4\frac{\sqrt{\ln\!\left(9-x\right)}\,dx}{\sqrt{\ln\!\left(9-x\right)}+\sqrt{\ln\!\left(x+3\right)}}$

All I know is that this equals 1 but I don't know how to get it. XD

Any input or suggestions would be appreciated!!

2. Originally Posted by Chris L T521
I was tutoring someone in Calculus today, and they had an extra credit assignment to do. These integrals appeared on past PUTNAM exams, and they can get help on these. However, I've been trying to work this one out and I (and all the other tutors I work with) couldn't figure it out.

So for the sake of me giving a hint or two to this person, I would like some guidance on how to tackle this beast:

$\int_2^4\frac{\sqrt{\ln\!\left(9-x\right)}\,dx}{\sqrt{\ln\!\left(9-x\right)}+\sqrt{\ln\!\left(x+3\right)}}$

All I know is that this equals 1 but I don't know how to get it. XD

Any input or suggestions would be appreciated!!
Consider

$
\int_2^4 \frac{\sqrt{\ln 9-x} + \sqrt{\ln x+3} }{\sqrt{\ln 9-x} + \sqrt{\ln x+3}}\,dx = \int_2^4 1 = 2
$

Expanding gives

$
\int_2^4 \frac{\sqrt{\ln 9-x} + \sqrt{\ln x+3} }{\sqrt{\ln 9-x} + \sqrt{\ln x+3}}\,dx = \int_2^4 \frac{\sqrt{\ln 9-x} }{\sqrt{\ln 9-x} + \sqrt{\ln x+3}}\,dx$
$+ \int_2^4 \frac{ \sqrt{\ln x+3} }{\sqrt{\ln 9-x} + \sqrt{\ln x+3}}\,dx$

Let $x = 6-u$ in the second integral

$\int_2^4 \frac{\sqrt{\ln 9-x} }{\sqrt{\ln 9-x} + \sqrt{\ln x+3}}\,dx$ $- \int_4^2 \frac{ \sqrt{\ln 9-u} }{\sqrt{\ln 3+u} + \sqrt{\ln 9-u}}\,dx$

$= 2 \int_2^4 \frac{\sqrt{\ln 9-x} }{\sqrt{\ln 9-x} + \sqrt{\ln x+3}}\,dx = 2$

$\int_2^4 \frac{\sqrt{\ln 9-x} }{\sqrt{\ln 9-x} + \sqrt{\ln x+3}}\,dx = 1$

3. Originally Posted by danny arrigo
Consider

$
\int_2^4 \frac{\sqrt{\ln 9-x} + \sqrt{\ln x+3} }{\sqrt{\ln 9-x} + \sqrt{\ln x+3}}\,dx = \int_2^4 1 = 2
$

Expanding gives

$
\int_2^4 \frac{\sqrt{\ln 9-x} + \sqrt{\ln x+3} }{\sqrt{\ln 9-x} + \sqrt{\ln x+3}}\,dx = \int_2^4 \frac{\sqrt{\ln 9-x} }{\sqrt{\ln 9-x} + \sqrt{\ln x+3}}\,dx$
$+ \int_2^4 \frac{ \sqrt{\ln x+3} }{\sqrt{\ln 9-x} + \sqrt{\ln x+3}}\,dx$

Let $x = 6-u$ in the second integral

$\int_2^4 \frac{\sqrt{\ln 9-x} }{\sqrt{\ln 9-x} + \sqrt{\ln x+3}}\,dx$ $- \int_4^2 \frac{ \sqrt{\ln 9-u} }{\sqrt{\ln 3+u} + \sqrt{\ln 9-u}}\,dx$

$= 2 \int_2^4 \frac{\sqrt{\ln 9-x} }{\sqrt{\ln 9-x} + \sqrt{\ln x+3}}\,dx = 2$

$\int_2^4 \frac{\sqrt{\ln 9-x} }{\sqrt{\ln 9-x} + \sqrt{\ln x+3}}\,dx = 1$

I would have never thought of that. Thank you very much!

4. It is very simple actually.Look at this standard property in definite integral
$
\int_a^bf(x)dx=\int_a^bf(a+b-x)dx
$

and the rest is obvious

Interesting corollary to above

$\int_a^b\frac{f(x)}{f(x)+f(a+b-x)}dx=\frac{b-a}{2}$

5. Originally Posted by pankaj
It is very simple actually.Look at this standard property in definite integral
$
\int_a^bf(x)dx=\int_a^bf(a+b-x)dx
$

and the rest is obvious

Interesting corollary to above

$\int_a^b\frac{f(x)}{f(x)+f(a+b-x)}dx=\frac{b-a}{2}$
The first part of this is in Stewart's Calculus book (problems plus).