How to prove the equality $\int^{na}_{ma}\frac{\ln(x-a)}{x^2+a^2}\,dx = \int^{\frac{a}{m}}_{\frac{a}{n}}\frac{\ln(x+a)}{x^2+a^2}\,dx$? How to prove the equality 
$$\int^{na}_{ma}\frac{\ln(x-a)}{x^2+a^2} \, dx= \int^{\frac{a}{m}}_{\frac{a}{n}}\frac{\ln(x+a)}{x^2+a^2}\, dx,$$
where $a, m, n$ are strictly positive numbers such that $mn=m+n+1$?
 A: $x=a^2/u$ should help you. For example, $dx/(x^2+a^2)=-du/(a^2+u^2)$. I believe the constraint comes into play when consisting the extra terms in the logarithm.
However, I'm not quite sure about the $\pm$ signs in the logarithm; are these correct?
This substitution is motivated by the observation that you go from $am\to a/m$ via $x\to a^2/x$.
A: My solution:
To note $I_{1}=\int^{na}_{ma}\frac{ln(x-a)}{x^{2}+a^{2}}dx$ and $I_{2}= \int^{\frac{a}{m}}_{\frac{a}{n}}\frac{ln(x+a)}{x^{2}+a^{2}}dx.$
Using the substitution $x=\frac{at+a^{2}}{t-a}$ we find $I_{1}=\frac{ln(2a^{2})}{2a}[\arctan (n)- \arctan (m)]$.
Using the substitution $x=\frac{-at+a^{2}}{t+a}$ we find $I_{2}=\frac{ln(2a^{2})}{2a}(\arctan \frac{1}{m}- \arctan \frac{1}{n})$.
Because for $m > 0$ and $n > 0$ is true equality 
$\arctan (m) + \arctan \frac{1}{m} = \frac{\pi}{2}= \arctan (n) + \arctan \frac{1}{n}$
then
$\arctan (n)-\arctan (m)= \arctan \frac{1}{m}-\arctan \frac{1}{n}$ 
and consequently
$I_{1}=I_{2}$.
For another solution see: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=296&t=536296
