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.
 A: For the first integral, we introduce a new parameter $a$.  Let $I(a)$ be the integral
$$I(a)=\int_0^\infty \frac{\log(a+x^4)}{1+x^2}\,dx \tag 1$$
We note that it can be shown that $I(0)=0$.  Our integral of interest is $I(1)$.
Proceeding, we differentiate $(1)$ to obtain 
$$\begin{align}
I'(a)&=\int_0^\infty\frac{1}{(1+x^2)(a+x^4)}\,dx\\\\
&=\frac{1}{2(a+1)}\int_{-\infty}^\infty\left(\frac{1}{x^2+1}+\frac{1-x^2}{x^4+a}\right)\,dx\\\\
&= \frac{2\pi i}{2(a+1)}\left(\frac{1}{2i}+\frac{1}{i\,2^{3/2}a^{3/4}}-\frac{1}{i\,2^{3/2}a^{1/4}}\right)\\\\
&=\frac{\pi}{2(1+a)}+\frac{\pi}{2^{3/2}(1+a)a^{3/4}}-\frac{\pi}{2^{3/2}(1+a)a^{1/4}}\tag 2
\end{align}$$
Integrating the right-hand side of $(2)$ reveals
$$\begin{align}
I(1)&=\frac{\pi}{2}\log(2)+\frac{\pi}{2^{3/2}}\int_0^1\left(\frac{1}{(1+x)x^{3/4}}-\frac{1}{(1+x)x^{1/4}}\right)\,dx\\\\
&=\frac{\pi}{2}\log(2)+\pi\sqrt 2\int_0^1\frac{1-x^2}{1+x^4}\,dx\\\\
&=\frac{\pi}{2}\log(2)+\pi\sqrt 2\left.\left(\frac{\log(x^2+\sqrt2 x+1)-\log(-x^2+\sqrt 2x-1)}{2^{3/2}}\right)\right|_{0}^{1}\\\\
&=\frac{\pi}{2}\log(2)+\frac{\pi}{2}\log\left(\frac{2+\sqrt 2}{2-\sqrt 2}\right)\\\\
&=\pi\log(2+\sqrt 2)
\end{align}$$
A: Express the 2nd integral in terms of the 1st
\begin{align}
&\int_{0}^{1}\frac{\arctan x}{x}\frac{1-x^{4}}{1+x^{4}}{d}x
 =\int_{0}^{1}\arctan x \ d\bigg(\frac12\ln \frac {x^2} {{1+x^4}} \bigg)\\
=&-\frac\pi8\ln2
+\frac12\int_0^1 \frac{\ln\frac{1+x^4}{x^2}}{1+x^2}\overset{x\to 1/x}{dx} 
 =-\frac\pi8\ln2
+\frac14 \int_0^\infty \frac{\ln(1+x^4)}{1+x^2}dx
\end{align}
A: 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
$$
\begin{aligned}
\int_0^{\infty} \frac{\ln \left(1+x^4\right)}{1+x^2} d x 
 & =\operatorname{Re}\left[2 \pi \ln \left(1+\frac{1+i}{\sqrt{2}}\right)\right] \\
& =2 \pi \operatorname{Re}\left[\ln \left(1+\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}\right)\right]\\
& =\pi \ln \left(1+\frac{1}{2}+\sqrt{2}+\frac{1}{2}\right) \\
& =\pi \ln (2+\sqrt{2})
\end{aligned}
$$
