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.

I hope it is not too late. Define
\begin{eqnarray}
I(a)=\int_0^\infty\frac{\log(1+ax^4)}{(1+x^2)^2}dx.
\end{eqnarray}
Then
\begin{eqnarray}
I'(a)&=&\int_0^\infty \frac{x^4}{(1+ax^4)(1+x^2)^2}dx\\
&=&\frac{1}{(1+a)^2}\int_0^\infty\left(-\frac{2}{1+x^2}+\frac{1+a}{(1+x^2)^2}+\frac{1-a+2ax^2}{1+a x^4}\right)dx\\
&=&\frac{1}{(1+a)^2}\left(-\pi+\frac{1}{4}(1+a)\pi+\frac{(1-a)\pi}{2\sqrt2a^{1/4}}+\frac{\pi a^{1/4}}{\sqrt2}\right)\\
&=&\frac{1}{4(1+a)^2}\left(a-3+\frac{\sqrt2(1-a)}{a^{1/4}}+2\sqrt2 a^{1/4}\right).
\end{eqnarray}
and hence
\begin{eqnarray}
I(1)&=&\int_0^1\frac{1}{4(1+a)^2}\left(a-3+\frac{\sqrt2(1-a)}{a^{1/4}}+2\sqrt2 a^{1/4}\right)da\\
&=&-\frac{\pi}{2}+\frac{1}{4}\log(6+4\sqrt2).
\end{eqnarray}
For the other integral, we can do the same thing to define
$$ J(a)=\int_0^\infty\frac{\log(1+ax^3)}{(1+x^2)^2}dx. $$
The calculation is similar and more complicated and here I omit the detail.
