Evaluate $\int_0^\infty \frac{\log(1+x^3)}{(1+x^2)^2}dx$ and $\int_0^\infty \frac{\log(1+x^4)}{(1+x^2)^2}dx$ Background: Evaluation of $\int_0^\infty \frac{\log(1+x^2)}{(1+x^2)^2}dx$
We can prove using the Beta-Function identity that
$$\int_0^\infty \frac{1}{(1+x^2)^\lambda}dx=\sqrt{\pi}\frac{\Gamma \left(\lambda-\frac{1}{2} \right)}{\Gamma(\lambda)} \quad \lambda>\frac{1}{2}$$
Differentiating the above equation with respect to $\lambda$, we obtain an expression involving the Digamma Function $\psi_0(z)$.
$$\int_0^\infty \frac{\log(1+x^2)}{(1+x^2)^\lambda}dx = \sqrt{\pi}\frac{\Gamma \left(\lambda-\frac{1}{2} \right)}{\Gamma(\lambda)} \left(\psi_0(\lambda)-\psi_0 \left( \lambda-\frac{1}{2}\right) \right)$$
Putting $\lambda=2$, we get
$$\int_0^\infty \frac{\log(1+x^2)}{(1+x^2)^2}dx = -\frac{\pi}{4}+\frac{\pi}{2}\log(2)$$

Question:
But, does anybody know how to evaluate $\displaystyle \int_0^\infty \frac{\log(1+x^3)}{(1+x^2)^2}dx$ and $\displaystyle \int_0^\infty \frac{\log(1+x^4)}{(1+x^2)^2}dx$?
Mathematica gives the values

*

*$\displaystyle \int_0^\infty \frac{\log(1+x^3)}{(1+x^2)^2}dx = -\frac{G}{6}+\pi \left(-\frac{3}{8}+\frac{1}{8}\log(2)+\frac{1}{3}\log \left(2+\sqrt{3} \right) \right)$


*$\displaystyle \int_0^\infty \frac{\log(1+x^4)}{(1+x^2)^2}dx = -\frac{\pi}{2}+\frac{\pi \log \left( 6+4\sqrt{2}\right)}{4}$
Here, $G$ denotes the Catalan's Constant.
Initially, my approach was to find closed forms for
$$\int_0^\infty \frac{1}{(1+x^2)^2(1+x^3)^\lambda}dx \ \ , \int_0^\infty \frac{1}{(1+x^2)^2(1+x^4)^\lambda}dx$$
and then differentiate them with respect to $\lambda$ but it didn't prove to be of any help.
Please help me prove these two results.

Letting $x\mapsto \frac{1}{x}$ reduces the power of the denominator and changes the integral into
$$
\begin{aligned}
\int_0^{\infty} \frac{\ln \left(1+x^n\right)}{\left(1+x^2\right)^2} d x & =\int_0^{\infty} \frac{x^2 \ln \left(1+x^n\right)}{\left(1+x^2\right)^2} d x-n \int_0^{\infty} \frac{x^2 \ln x}{\left(1+x^2\right)^2} d x \\
& =\int_0^{\infty} \frac{1+x^2-1}{\left(1+x^2\right)^2} \ln \left(1+x^n\right) d x-\frac{n \pi}{4} \\
& =\frac{1}{2} \int_0^{\infty} \frac{\ln \left(1+x^n\right)}{1+x^2} d x-\frac{n \pi}{8} .
\end{aligned}
$$
Hence
$$
\begin{aligned}
\int_0^{\infty} \frac{\ln \left(1+x^3\right)}{\left(1+x^2\right)^2} d x & =\frac{1}{2} \int_0^{\infty} \frac{\ln \left(1+x^3\right)}{1+x^2} d x-\frac{3 \pi}{8} \\
& =\frac{1}{2}(-\frac{G}{3}+\frac{\pi}{4} \ln 2 +\frac{2 \pi}{3}\ln(2+\sqrt{3})-\frac{3 \pi}{8} \\
& =-\frac{G}{6}+\pi\left(-\frac{3}{8}+\frac{1}{8} \ln 2+\frac{1}{3} \ln (2+\sqrt{3})\right)
\end{aligned}
$$
The second last line comes from my post.

$$
\begin{aligned}
I_4 & =\frac{1}{2} \int_0^{\infty} \frac{\ln \left(1+x^4\right)}{1+x^2}-\frac{\pi}{2} \\
& =\frac{\pi \ln (6+4 \sqrt{2})} {4}-\frac{\pi}{2}
\end{aligned}
$$
The second last line comes from my post.
