I am dealing with this sum of two divergent integrals
$$-\frac{1}{2}\int _0^{1}\frac{\left(\frac{1-t}{1+t}\right)\arctan \left(t\right)}{t\left(\ln t\right)^2}dt-\frac{\pi }{4}\int _0^{1}\frac{1}{\ln \left(t\right)\left(1+t\right)^2}dt$$
This is ugly but although they diverge seperately, their sum converges to $0.1746809...$
Can we find a closed form in terms of known constants for this number?
I've tried getting like denominators by setting $t\to e^{-\sqrt{|\ln(t)|}}$ in the second integral.
Which gave $$-\frac{\pi }{4}\int _0^{1}\frac{\frac{2\left(\ln t\right)e^{\left(\ln t\right)^2}}{\left(1+e^{\left(\ln t\right)^2}\right)^2}}{t\left(\ln t\right)^2}dt$$
However I realized that the expression has to be written in terms of the limit $m\to 1$, because although the upper bound in this integral is $1$, both integrals separately diverge at different rates as their upper bounds go to $1$. So we'd have to write the expression as
$$\lim_{m\to 1}\left(-\frac{1}{2}\int _0^{m}\frac{\left(\frac{1-t}{1+t}\right)\arctan t}{t\left(\ln t\right)^2}dt-\frac{\pi }{4}\int _0^{e^{-\sqrt{|\ln m|}}}\frac{\frac{2\left(\ln t\right)e^{\left(\ln t\right)^2}}{\left(1+e^{\left(\ln t\right)^2}\right)^2}}{t\left(\ln t\right)^2}dt\right)$$
Thus we can't combine these integrals and this doesn't work.
Due to Jack's comment I will rewrite the integral as $$\int _0^{1}\frac{dt}{\ln t}\left(\frac{\pi }{2\left(1+t\right)^2}+\frac{1-t}{1+t}\frac{\arctan t}{t\ln t}\right)$$ which is just twice the original value.