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

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


*$\displaystyle \int_0^\infty \frac{\log(1+x^4)}{(1+x^2)^2}dx = -\frac{\pi}{2}+\frac{\pi \log \left( 6+4\sqrt{2}\right)}{4}$
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.

Another approach for evaluating the second integral using contour integration that avoids having to deform the contour around branch cuts is to consider $$ \displaystyle f(z) = \frac{\log(z+ e^{i \pi /4})}{(1+z^{2})^{2}}$$ and integrate around a contour that consists of the line segment $[-R,R]$ and the upper half of $|z|=R$.
Then  letting $R \to \infty$, 
$$ \begin{align} &\int_{-\infty}^{0} \frac{\log(x+e^{i \pi /4})}{(1+x^{2})^{2}} \ dx + \int_{0}^{\infty} \frac{\log(x+e^{i \pi /4})}{(1+x^{2})^{2}} \ dx \\ &= 2 \pi i \ \text{Res}[f(z),i] \\ &= 2 \pi i \lim_{z \to i} \frac{d}{dz} \frac{\log(z+e^{i \pi /4})}{(z+i)^{2}} \\ &=2 \pi i \lim_{z \to i} \left(\frac{1}{(z+e^{i \pi /4})(z+i)^{2}} - \frac{2 \log(z+e^{i \pi /4})}{(z+i)^{3}} \right) \\ &= 2 \pi i \left(- \frac{1}{4} \frac{\sqrt{2}}{1+i(1+\sqrt{2})} + \frac{\log|i+e^{i \pi /4}| + i \arg (i +e^{i \pi/4}) }{4i}\right)\\ &= 2 \pi i \left(\frac{1-\sqrt{2}+i}{8} + \frac{\frac{1}{2} \log (2+\sqrt{2})+ i \frac{3 \pi}{8}}{4i} \right) \\ &= \frac{\pi}{4} \Big(\log(2+\sqrt{2})-1 \Big) + \frac{i\pi}{4} \left(1-\sqrt{2}+\frac{3 \pi}{4} \right) . \end{align}$$
But notice that $$ \begin{align} &\text{Re} \left( \int_{-\infty}^{0} \frac{\log(z+e^{i \pi /4})}{(1+x^{2})^{2}} \ dx + \int_{0}^{\infty} \frac{\log(z+e^{i \pi /4})}{(1+x^{2})^{2}} \ dx \right) \\ &= \text{Re} \left(\int_{0}^{\infty} \frac{\log(-u + e^{i \pi /4})}{(1+u^{2})^{2}} \ du +  \int_{0}^{\infty} \frac{\log(z+e^{i \pi /4})}{(1+x^{2})^{2}} \ dx \right) \\ &= \int_{0}^{\infty} \frac{\log|-u + e^{i \pi /4}|}{(1+u^{2})^{2}} \ du + \int_{0}^{\infty} \frac{\log |x+e^{i \pi /4}|}{(1+x^{2})^{2}} \ dx\\ &= \int_{0}^{\infty} \frac{\frac{1}{2} \log(x^{2}-\sqrt{2}x+1) + \frac{1}{2} \log(x^{2}+\sqrt{2}x+1)}{(1+x^{2})^{2}} \ dx \\ &= \frac{1}{2} \int_{0}^{\infty} \frac{\log(1+x^{4})}{(1+x^{2})^{2}} \ dx. \end{align}$$
Therefore,
$$ \begin{align} \int_{0}^{\infty} \frac{\log(1+x^{4})}{(1+x^{2})^{2}} \ dx &= \frac{\pi}{2} \Big(\log(2+\sqrt{2}) -1 \Big) \\ &= \frac{\pi}{2} \Big(\frac{1}{2} \log \big((2+\sqrt{2})^{2} \big) -1 \Big)  \\ &= \frac{\pi}{2} \Big(\frac{\log(6+4\sqrt{2})}{2} -1\Big) \\ &= - \frac{\pi}{2} + \frac{\pi \log(6+4\sqrt{2})}{4}. \end{align}$$
