TITLE: Integral $\int_0^{1}\frac{(1+x^2)\ln(1+x^4)}{1+x^4}dx$
QUESTION [6 upvotes]: Find: $$I=\int_0^{1}\frac{(1+x^2)\ln(1+x^4)}{1+x^4}\,dx$$

I need to use the definition of do Trigamma function but I don't see how.
My try was as following : 
Let $y=x^4$ then $dy=4x^{3}\,dx$ so $\;dx=\frac{dy}{4y^{\frac{3}{4}}}$.
$$I=\int_0^{1}\frac{(1+x^{\frac{3}{4}})\ln(1+x)}{x^{\frac{3}{4}}+x^{\frac{7}{4}}}dx$$
Use : $\ln(1+x)=-\sum_{n=1}^{\infty}\frac{(-1)^{n}x^n}{n}$ 
We obtain 
$$J=\int_0^{1}\frac{x^n+x^{(3+4n)/4}}{x^{3/4}+x^{7/4}}dx$$
Then I use definition of digamma function
$$\int_0^{1}\frac{x^{a-1}\ln x}{1+x}\,dx=\beta'(a)$$
Where:  $$2\beta(a)=\psi\left(\frac{1+a}{2}\right)-\psi\left(\frac{a}{2}\right)$$
But I find sum related with digamma function.
So I don't have other ideas to approach it!

REPLY [5 votes]: (Too long for a comment.)
Zacky gave the answer:
$$\small \int_0^1 \frac{(1+x^2)\ln(1+x^4)}{1+x^4}dx=\frac{3\pi}{2\sqrt 2}\ln 2-\frac1{32}\left(\psi_1\left(\frac{1}{8}\right)+\psi_1\left(\frac{3}{8}\right)-\psi_1\left(\frac{5}{8}\right)-\psi_1\left(\frac{7}{8}\right)\right)$$
After my post on Catalan's constant $K = \rm{Cl}_2\big(\tfrac\pi2\big)$ and Gieseking's constant $\kappa =\rm{Cl}_2\big(\tfrac\pi3\big)$, I recognized the four polygammas as involving Catalan's quartic counterpart $\rm{Cl}_2\big(\color{red}{\tfrac\pi4}\big)$, namely,
$$-\small\frac1{32} \left(\psi_1\left(\frac{1}{8}\right)+\psi_1\left(\frac{3}{8}\right)-\psi_1\left(\frac{5}{8}\right)-\psi_1\left(\frac{7}{8}\right)\right) = \frac1{\sqrt2}\Big(K-4\,\rm{Cl}_2\big(\color{red}{\tfrac\pi4}\big)\Big)$$
Hence we have the simple closed-form,
$$\int_0^1 \frac{(1+x^2)\ln(1+x^4)}{1+x^4}dx=\frac{3\pi}{2\sqrt2}\ln 2+ \frac K{\sqrt2} -\frac{4\rm{Cl}_2\big(\color{red}{\tfrac\pi4}\big)}{\sqrt2}$$
where $\rm{Cl}_2\big(\tfrac\pi4\big) = \Im\, \rm{Li}_2(e^{2\pi i/8})$.