<s>[INST] 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.
 [/INST] Alternatively, with
$ \frac{1+t^2}{1-t^2+t^4}= \frac{1}{1+t^2} +\frac{3t^2}{1+t^6}$
\begin{align}
&\int_0^1 \frac{(1+t^2)\ln t}{1-t^2+t^4}dt
=\int_0^1 \underset{=-G}{\frac{\ln t}{1+t^2}}dt
+ \int_0^1 \underset{=-\frac13G}{\frac{3t^2\ln t}{1+t^6} \ \overset{t^3\to t}{dt}}=-\frac43G
\end{align}
</s>