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.

The 2nd evaluation can be obtained from the residue theorem:


*

*Using parity, write the integral as $\displaystyle \frac12\int_{-\infty}^{\infty}\frac{\ln(1+x^4)\,dx}{(1+x^2)^2}$.

*Interpret this as a complex integral and pull the integration contour to, say, $i\infty$. The result will be given by the residue at 2nd order pole at $x=i$ and two integrals of the logarithm jump (equal to $2\pi i$) over the branch cuts emanating from $x=e^{i\pi/4}$ and $x=e^{3i\pi /4}$ in the radial directions.

*For the first contribution, we have
$$2\pi i \cdot\mathrm{res}_{x=i}\frac{\ln(1+x^4)}{(1+x^2)^2}=\frac{\pi}{2}\left(\ln2 -2\right).$$

*The integral over the branch cut $(e^{i\pi/4},e^{i\pi/4}\infty)$ is
$$2\pi i \int_{e^{i\pi/4}}^{e^{i\pi/4}\infty}\frac{dx}{(1+x^2)^2}=\frac{\pi i}{2}\left(\pi-\sqrt{2}-2\arctan e^{i\pi/4}\right),$$
and, similarly, for the second branch cut $(e^{3i\pi/4},e^{3i\pi/4}\infty)$ we find
$$2\pi i \int_{e^{3i\pi/4}}^{e^{3i\pi/4}\infty}\frac{dx}{(1+x^2)^2}=\frac{\pi i}{2}\left(\sqrt{2}-\pi-2\arctan e^{3i\pi/4}\right).$$

*Combining everything, one obtains the answer:
\begin{align}
\frac12\left\{\frac{\pi}{2}\left(\ln2 -2\right)+\pi\, \mathrm{arccoth}\sqrt{2}\right\}=
-\frac{\pi}{2}+\frac{\pi}{4}\ln(6+4\sqrt{2}).
\end{align}
