Prove that $ \int_0^1\frac{x\ln(x)}{1+x^2+x^4}dx=\frac{1}{36}\Big(\psi^{(1)}(2/3)-\psi^{(1)}(1/3)\Big) $ I am having trouble with the following integral

Prove that $$ \int_0^1\frac{x\ln(x)}{1+x^2+x^4}dx=\frac{1}{36}\Big(\psi^{(1)}(2/3)-\psi^{(1)}(1/3)\Big)$$

$$I=\int_0^1\frac{x\ln(x)}{1+x^2+x^4}dx=\int_0^1\frac{\ln(u^2)}{2(1+u+u^2)}du=\int_0^1\frac{\ln(u)}{(1+u+u^2)}du$$
let $x^2=u\rightarrow \frac{du}{dx}=2x$
How does one proceed from here? Is my approach correct? Thank you for your time
 A: Firstly, note that after your substitution it should be
$$I=\int_0^1\frac{x\ln(x)}{1+x^2+x^4}~dx=\color{red}{\frac{1}{4}}\int_0^1\frac{\ln(u)}{1+u+u^2}~du.$$
To evaluate the latter integral, the geometric series shows that
$$\begin{align*} \int_0^1\frac{\ln(u)}{1+u+u^2}~du&=\int_0^1 \frac{(1-u)\ln(u)}{1-u^3}~du\\&=\int_0^1 \sum_{k=0}^{\infty} (1-u)\ln(u)u^{3k}~du\\&=\sum_{k=0}^{\infty} \left[\int_0^1 u^{3k}\ln(u)~du-\int_0^1 u^{3k+1}\ln(u)~du\right]. \end{align*}$$
By differentiating the integral $\int_0^1 x^{\alpha}~dx$ with respect to $\alpha\in \mathbb{R}\setminus \{-1\}$, one obtains that
$$\int_0^1 x^{\alpha}\ln(x)~dx=-\frac{1}{(\alpha+1)^2}.$$
Therefore, one has that
$$\begin{align*} \int_0^1\frac{\ln(u)}{1+u+u^2}~du&=\sum_{k=0}^{\infty} \left[\frac{1}{(3k+2)^2}-\frac{1}{(3k+1)^2}\right]\\&=\frac{1}{9}\sum_{k=0}^{\infty} \left[\frac{1}{(k+2/3)^2}-\frac{1}{(k+1/3)^2}\right]\\&=\frac{1}{9}(\psi^{(1)}(2/3)-\psi^{(1)}(1/3)), \end{align*}$$
where we have used the series representation of the trigamma function
$$\psi^{(1)}(z)=\sum_{k=0}^{\infty} \frac{1}{(z+k)^2}. \tag{1}$$
If this is not your definition of the trigamma function (if you define $\psi^{(1)}(z):=\frac{d^2}{dz^2}\ln(\Gamma(z))$), then you can prove $(1)$ using the Weierstrass's definition of the $\Gamma$ function:
$$\Gamma(z)=\frac{e^{-\gamma z}}{z}\prod_{n=1}^{\infty}  \left(1+\frac{z}{n}\right)^{-1}e^{z/n}.$$
A: Alternatively
\begin{align} \int_0^1\frac{x\ln x}{1+x^2+x^4}dx
= &\int_0^1\frac{x\ln x}{(e^{i\frac\pi3}+x^2)(e^{-i\frac\pi3}+x^2)}dx \\
=& \>\frac1{\sin\frac\pi3 }\>\Im \int_0^1 \frac{e^{i\frac\pi3}x\ln x}{1+ e^{i\frac\pi3}x^2 }dx
\overset{x^2\to x} =\frac{1}{2\sqrt3}\Im \text{Li}_2(- e^{i\frac\pi3})
\end{align}
