This is related to this question.
Let $$f(t) = \int_0^1 \frac{x \ln(x)}{(x^2+t)^2} dx$$
so that we want $f(1)$. (The answer turns out to be $-\frac{1}{4} \log(2)$.)
Then $$f'(t) = -2 \int_0^1 \frac{x \ln(x)}{(x^2+t)^3} dx$$
and in general we get inductively $$f^{(n)}(t) = (-1)^n (n+1)! \int_0^1 \frac{x \ln(x)}{(x^2+t)^{n+1}} dx$$
so $$\frac{f^{(n)}(t)}{n!} = (-1)^n (n+1) \int_0^1 \frac{x \ln(x)}{(x^2+t)^{n+1}} dx$$
and hence $$\frac{f^{(n)}(0)}{n!} = (-1)^n (n + 1) \int_0^1 \frac{\log(x)}{x^{2n+1}} dx$$
This integral diverges unless $n < 0$, but in those conditions it comes to $-\frac{1}{4n^2}$. Let's carry on regardless.
Now by Taylor expansion, $$f(1) = f(0) + f'(0) + \frac{f''(0)}{2} + \dots$$
Ignoring convergence issues, $$f(1) = f(0) - \sum_{n=1}^{\infty} \frac{1}{4n^2} (-1)^n (n + 1)$$
And that sum is $$\frac{1}{48} (\pi^2 + 12 \log(2))$$
so we obtain $$f(1) = f(0) - \frac{\pi^2}{48} - \frac{1}{4} \log(2)$$
If we can manipulate $\int_0^1 \frac{\log(x)}{x} dx$ to come out as $\frac{\pi^2}{48}$, then we've arrived at the answer by completely spurious means.
Is there any way to rigorise this answer? I suspect there might be, because the final answer differs by a rather easy constant from the actual answer. My other idea is that perhaps the calculation of the final sum requires calculating my original integral $f(1)$, and so we've done something circular and found the right result because we assumed it.
