Integral $\int_0^1 \frac{(1+t^2)}{1-t^2+t^4}\ln(t)dt$ $$I=\int_0^1 \frac{(1+t^2)}{1-t^2+t^4}\ln(t)dt=-\frac{4}{3}G  $$ G is Catalan's constant.
$$I=\int_0^1 \frac{(1+t^2)^2}{(1+t^2)(1-t^2+t^4)}\ln(t)dt=\int_0^1 \frac{(1+t^2)^2}{(1+t^6)}\ln(t)dt$$
Do series expansion:
$$I=\sum_{n=0}^\infty (-1)^n \int_0^1(1+2t^2+t^4)t^{6n}\ln(t)dt$$
Integrate term by term:
$$I=-\sum_{n=0}^\infty (-1)^n\left( \frac{1}{(6n+1)^2}+\frac{2}{(6n+3)^2}+\frac{1}{(6n+5)^2}  \right)$$
Re-organize them:
$$I=-\sum_{n=0}^\infty (-1)^n\left( \frac{1}{(6n+1)^2}-\frac{1}{(6n+3)^2}+\frac{1}{(6n+5)^2}\right)-\sum_{n=0}^\infty (-1)^n \frac{3}{(6n+3)^2} $$
The first part is Catalan's constant:
$$I=-G-\sum_{n=0}^\infty (-1)^n \frac{3}{(6n+3)^2}=-G-\frac{3}{9}G=-\frac{4}{3}G$$
Done.

You can arrive to the desired result starting at
$$I=-\sum_{n=0}^\infty (-1)^n\left( \frac{1}{(6n+1)^2}+\frac{2}{(6n+3)^2}+\frac{1}{(6n+5)^2}  \right)$$
Using the Hurwitz zeta function
$$S_a=\sum_{n=0}^\infty \frac{ (-1)^n}{(6n+a)^2}=\frac{1}{144} \left(\zeta \left(2,\frac{a}{12}\right)-\zeta  \left(2,\frac{a+6}{12}\right)\right)$$
$$S_1+2S_3+S_5=S_1+\frac{2}{9}C+S_5=$$ $$\frac{2}{9}C+\frac{1}{144}\left(\zeta \left(2,\frac{1}{12}\right)+\zeta \left(2,\frac{5}{12}\right)-\zeta
   \left(2,\frac{7}{12}\right)-\zeta \left(2,\frac{11}{12}\right)\right)$$ and use
$$\zeta \left(2,\frac{1}{12}\right)+\zeta \left(2,\frac{5}{12}\right)-\zeta
   \left(2,\frac{7}{12}\right)-\zeta \left(2,\frac{11}{12}\right)=160 C$$  You could do the same using the first derivative of the digamma function.
Edit
We could do the same without series expansion writing
$$\frac{(1+t^2)}{1-t^2+t^4}=\frac 1{a-b}\left(\frac{a+1}{t^2-a}-\frac{b+1}{t^2-b} \right)$$ with $a=\frac{1+i \sqrt{3}}{2}$ and $b=\frac{1-i \sqrt{3}}{2}$ and use, for a complex $c$,
$$J(c)=\int_0^1 \frac{\log(t)}{t^2-c}\,dt=\frac{1}{4 \sqrt{c}}\left(4
   \text{Li}_2\left(\frac{1}{\sqrt{c}}\right)-\text{Li}_2\left(\frac{1}{c}\right)\right)$$
