Integral could be worth generalising

An alternate way of approaching the problem:

$I = \int_{0}^{\infty} \frac{\ln{x}}{N^2 + x^2}dx$

Let: $x = N\tan{\theta}$ $dx = N\sec^2{\theta}d\theta$

When $x = 0$ , $\theta = 0$ and when $x = \infty$ , $\theta = \pi/2$ . This gives:

$I = \int_{0}^{\frac{\pi}{2}}\frac{\ln(N\tan{\theta})}{N^2 + N^2\tan^2{\theta}}N\sec^2{\theta}d\theta$

$I = \int_{0}^{\frac{\pi}{2}}\frac{\ln(N\tan{\theta})}{N^2\sec^2{\theta}}N\sec^2{\theta}d\theta$

$I = \frac{1}{N}\int_{0}^{\frac{\pi}{2}}\ln(N\tan{\theta})d\theta$

Knowing that:

$\int_{a}^{b}f(x)dx = \int_{a}^{b}f(a+b-x)dx$

$I = \frac{1}{N}\int_{0}^{\frac{\pi}{2}}\ln(N\cot{\theta})d\theta$

Therefore:

$2I = \frac{1}{N}\int_{0}^{\frac{\pi}{2}} \left(\ln(N\tan{\theta}) + \ln(N\cot{\theta})\right)d\theta$

$I = \frac{1}{2N}\int_{0}^{\frac{\pi}{2}}\ln(N^2)d\theta$

$I = \frac{\ln(N)}{N}\int_{0}^{\frac{\pi}{2}}d\theta$

$\boxed{I = \frac{\pi\ln(N)}{2N}}$

Taking $N = 25$ gives:

$I = \frac{\pi\ln(25)}{50} = \frac{\pi\ln(5)}{25}$

This leads to: $m = 25$ and $n = 5$ and therefore the final answer is:

$\boxed{m+n = 30}$

The substitution $x = y^{-1}$ gives us that $\int_0^\infty \frac{\ln x}{1 + x^2}\,dx \; = \; -\int_0^\infty \frac{\ln y}{1 + y^2}\,dy \; = \; 0$ and now the sibstitution $x = Nu$ gives $\int_0^\infty \frac{\ln x}{N^2 + x^2}\,dx \; = \; \frac{1}{N}\int_0^\infty \frac{\ln N + \ln u}{1 + u^2}\,du \; = \; \frac{1}{N}\ln N \int_0^\infty \frac{du}{1+u^2} \; = \; \frac{1}{2N}\pi\ln N$ With $N=25$ we see that $I = \tfrac{1}{50}\pi\ln25 = \tfrac{1}{25}\pi\ln5$ , making the answer $25+5=\boxed{30}$ .