How to evaluate $\int_{0}^{\infty }\frac{\ln( 1+x^{4} )}{1+x^{2}}{d}x~,~\int_{0}^{1}\frac{\arctan x}{x}\frac{1-x^{4}}{1+x^{4}}{d}x$ How to evaluate these two integrals below
$$\int_{0}^{\infty }\frac{\ln\left ( 1+x^{4} \right )}{1+x^{2}}\mathrm{d}x$$
$$\int_{0}^{1}\frac{\arctan x}{x}\frac{1-x^{4}}{1+x^{4}}\mathrm{d}x$$
For the first one, I tried to use
$$\mathcal{I}'(s)=\int_{0}^{\infty }\frac{x^4}{(1+sx^4)(1+x^{2})}\mathrm{d}x$$
but it seems hard to solve.

Using the identity $\ln \left(x^2+y^2\right)=2 \operatorname{Re}(\ln (x+y i))$ to reduce the power $4$ of $x$ in the numerator makes the life easier.
$$
\begin{aligned}
\int_0^{\infty} \frac{\ln \left(1+x^4\right)}{1+x^2} d x 
= & \frac{1}{2} \int_{-\infty}^{\infty} \frac{\ln \left(1+x^4\right)}{1+x^2} d x \\
= & \operatorname{Re}\left(\int_{-\infty}^{\infty} \frac{\ln \left(x^2+i\right)}{1+x^2}\right)\\=&\operatorname{Re} \int_{-\infty}^{\infty} \frac{\ln \left(x^2+\left(\frac{1+i}{\sqrt{2}}\right)^2\right)}{1+x^2}
\end{aligned}
$$
Using the particular case  $\int_{-\infty}^{\infty} \frac{\ln \left(x^2+a^2\right)}{1+x^2} d x=2 \pi \ln (1+a)$ in the post, we have
$$
\begin{aligned}
\int_0^{\infty} \frac{\ln \left(1+x^4\right)}{1+x^2} d x 
 & =\operatorname{Re}\left[2 \pi \ln \left(1+\frac{1+i}{\sqrt{2}}\right)\right] \\
& =2 \pi \operatorname{Re}\left[\ln \left(1+\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}\right)\right]\\
& =\pi \ln \left(1+\frac{1}{2}+\sqrt{2}+\frac{1}{2}\right) \\
& =\pi \ln (2+\sqrt{2})
\end{aligned}
$$
