$$\int_{0}^{1} \frac{\ln(1+x^2)\arctan(x^2)dx}{1+x^2}$$
I couldn't think of much except taking $x=\tan(t)$ which reduced the integral to $$\int_{0}^{\pi/4}\ln(\sec^2(t))\arctan(\tan^2(t))dt$$
Which still doesn't looks good to me.
Taking $\arctan(x^2)=t$ doesn't help much either. Neither does $\ln(1+x^2)=t$