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.
 A: 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}
A: 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.
A: 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 $c$,
$$J(c)=\int_0^1 \frac{\log(t)}{t^2-c}\,dt=\frac{1}{4 \sqrt{c}}\left(4
   \text{Li}_2\left(\frac{1}{\sqrt{c}}\right)-\text{Li}_2\left(\frac{1}{c}\right)\right)$$
