x Log[1+n^2]/n^2 + c

unless you really want integral log(1+x^2) / x^2 dx

Then we can write:

log(1+x^2) = x^2 - x^4/2 + x^6/3 - ... + (-1)^(n+1) x^(2n)/n + ...

for |x|<1.

Then:

integral log(1+x^2) / x^2 dx = sum_{n=1 to infnty} integral (-1)^{n+1}x^{2n-1}/n dx

.................= sum_{n=1 to infnty} (-1)^{n+1}x^{2n}/(n (2n-1)

RonL