Prove $\int_0^1 \frac{t^2-1}{(t^2+1)\log t}dt = 2\log\left( \frac{2\Gamma \left( \frac{5}{4}\right)}{\Gamma\left( \frac{3}{4}\right)}\right)$

Let's introduce the parameter $\alpha$, and then differentiate with respect to $\alpha$ that yields $$I(\alpha)=\int_0^1 \frac{t^\alpha-1}{(t^2+1)\ln t}dt $$ $$I'(\alpha)=\int_0^1 \frac{t^{\alpha}}{(t^2+1)}dt=\frac{1}{4} \left(-\psi_0\left(\frac{1 + \alpha}{4}\right) + \psi_0\left(\frac{3 + \alpha}{4}\right)\right) $$ Then $$I(\alpha)=\frac{1}{4} \int\left(-\psi_0\left(\frac{1 + \alpha}{4}\right) + \psi_0\left(\frac{3 + \alpha}{4}\right)\right) d\alpha= $$ $$I(\alpha)=\left(\ln \Gamma \left(\frac{3 + \alpha}{4}\right)- \ln \Gamma \left(\frac{1 + \alpha}{4}\right)\right)+C\tag1$$ If letting $\alpha=2$, then $$I(2)=\ln \left(\frac{\Gamma \left(\frac{5}{4}\right)}{\Gamma \left(\frac{3}{4}\right)}\right)+C$$ On the other hand, by letting $\alpha=0$ in $(1)$ we get $$C=\ln \left(\frac{\Gamma \left(\frac{1}{4}\right)}{\Gamma \left(\frac{3}{4}\right)}\right)$$ Thus $$\int_0^1 \frac{t^2-1}{(t^2+1)\ln t}dt=\ln \left(\frac{\Gamma \left(\frac{5}{4}\right)\Gamma \left(\frac{1}{4}\right)}{\Gamma^2 \left(\frac{3}{4}\right)}\right) $$

$$ \begin{align} \int_0^1\frac{t+1}{t^2+1}\frac{t-1}{\log(t)}\,\mathrm{d}t &=\int_0^1\frac{t+1}{t^2+1}\int_0^1t^x\,\mathrm{d}x\,\mathrm{d}t\\ &=\int_0^1\int_0^1\frac{t^{x+1}+t^x}{t^2+1}\,\mathrm{d}t\,\mathrm{d}x\\ &=\int_0^1\left(\frac1{x+1}+\frac1{x+2}-\frac1{x+3}-\frac1{x+4}+\dots\right)\,\mathrm{d}x\\ &=\left(\log\left(\frac21\right)+\log\left(\frac32\right)\right) -\left(\log\left(\frac43\right)+\log\left(\frac54\right)\right)+\dots\\ &=\log\left(\frac31\right)-\log\left(\frac53\right)+\log\left(\frac75\right)-\log\left(\frac97\right)+\log\left(\frac{11}9\right)-\dots\\ &=\log\left(\frac31\cdot\frac35\cdot\frac75\cdot\frac79\cdot\frac{11}9\cdots\right)\\ &=\lim_{n\to\infty}\log\left(\frac{\Gamma\left(\frac54\right)^2}{\Gamma\left(\frac34\right)^2} \frac{\Gamma\left(\frac{4n+3}4\right)^2}{\Gamma\left(\frac{4n+5}4\right)^2}(4n+3)\right)\\ &=2\log\left(2\frac{\Gamma\left(\frac54\right)}{\Gamma\left(\frac34\right)}\right) \end{align} $$ The last equality is due to Gautschi's inequality.