<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] Addendum: This is in response to the original question asked by the OP, which was essentially why the below is true:
$$\sum_{n=0}^\infty (-1)^n\left( \frac{1}{(6n+1)^2}+\frac{1}{(6n+5)^2}\right)=\frac{10}{9}G$$

So the obvious way to split this up would be into the sums
$$\newcommand{\AA}{\mathcal{A}}
\newcommand{\BB}{\mathcal{B}}
\begin{align*}
\AA &:= \sum_{n=0}^{\infty} \frac{(-1)^n}{(6n+1)^2} = 1 - \frac 1{7^2} + \frac{1}{13^2} - \frac{1}{19^2} \cdots \\
\BB &:= \sum_{n=0}^\infty \frac{(-1)^n}{(6n+5)^2} = \frac{1}{5^2} -\frac{1}{11^2} + \frac{1}{17^2} \cdots
\end{align*}$$
Recall, by definition,
$$G := \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^2} = 1 - \frac{1}{3^2} + \frac{1}{5^2} - \frac{1}{7^2} \cdots$$
Notice that
$$\AA + \BB = G - \frac{1}{3^2} + \frac{1}{9^2} - \frac{1}{15^2} + \frac{1}{21^2} \cdots$$
Multiply the bottom of each term in $G$'s summation by $-3^2$. Then
$$-\frac 1 9 G =  -\frac{1}{3^2} + \frac{1}{3^2} \frac{1}{3^2} - \frac{1}{3^2} \frac{1}{5^2} + \frac{1}{3^2} \frac{1}{7^2} \cdots$$
This is, once you simplify, precisely the missing terms. You can also see this as
$$\AA + \BB = G + \sum_{n=0}^\infty \frac{(-1)^n}{(6n+3)^2}$$
and
$$\sum_{n=0}^\infty \frac{(-1)^n}{(6n+3)^2}  
= \sum_{n=0}^\infty \frac{(-1)^n}{3^2 (2n+1)^2} 
= \frac 1 9 \sum_{n=0}^\infty \frac{(-1)^n}{ (2n+1)^2} 
= \frac 1 9 G$$
Hence,
$$\AA + \BB = G + \frac 1 9 G = \frac{10}{9} G$$
as desired.
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