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.
 A: 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.
A: 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,
$$ \begin{align} \int_{0}^{\infty} \frac{\log(1+x^{4})}{(1+x^{2})^{2}} \ dx &= \frac{\pi}{2} \Big(\log(2+\sqrt{2}) -1 \Big) \\ &= \frac{\pi}{2} \Big(\frac{1}{2} \log \big((2+\sqrt{2})^{2} \big) -1 \Big)  \\ &= \frac{\pi}{2} \Big(\frac{\log(6+4\sqrt{2})}{2} -1\Big) \\ &= - \frac{\pi}{2} + \frac{\pi \log(6+4\sqrt{2})}{4}. \end{align}$$
A: The generalized results for the even and odd cases of $$I_n=\int_0^\infty \frac{\log(1+x^n)}{(1+x^2)^2}dx
$$
are respectively as follows
\begin{align}
I_{2m} =& -\frac{m\pi}4+\frac{m\pi}2\ln2+\pi\sum_{k=1}^{[\frac m2]}\ln \cos\frac{(m-2k+1)\pi}{4m}\\
I_{2m+1} =& -\frac{(2m+1)\pi}8+\frac{(4m+1)\pi}8\ln2+\frac{(-1)^m G}{2(2m+1)}\\
&\ +\frac\pi2\sum_{k=0}^{m-1}\left[\ln \cos\frac{(2k+1)\pi}{4(2m+1)}+\frac{(-1)^{m+k}(2k+1)}{2(2m+1)}\ln\tan\frac{(2k+1)\pi}{4(2m+1)}
\right]\\
\end{align}
Specifically
\begin{align}
 \int_0^\infty \frac{\log(1+x)}{(1+x^2)^2}dx
=& -\frac\pi8+\frac\pi8\ln2 +\frac12G\\
\int_0^\infty \frac{\log(1+x^2)}{(1+x^2)^2}dx
=& -\frac\pi4+\frac\pi2\ln2 \\
\int_0^\infty \frac{\log(1+x^3)}{(1+x^2)^2}dx
=& -\frac{3\pi}8+\frac{\pi}8\ln2
+\frac\pi3\ln(2+\sqrt3) -\frac16G\\
\int_0^\infty \frac{\log(1+x^4)}{(1+x^2)^2}dx
=& -\frac\pi2+\frac\pi2\ln(2+\sqrt2)\\
 \int_0^\infty \frac{\log(1+x^5)}{(1+x^2)^2}dx
=& -\frac{5\pi}8-\frac{3\pi}8\ln2+\frac\pi2\ln\left(1+\sqrt5+\sqrt{2(5+\sqrt5)}\right)\\
&\  +\frac\pi{20}\ln\tan\frac\pi{20}-\frac{3\pi}{20}\ln\tan\frac{3\pi}{20} +\frac1{10}G\\
 \int_0^\infty \frac{\log(1+x^6)}{(1+x^2)^2}dx
=& -\frac{3\pi}4+\frac\pi2\ln6\\
 \int_0^\infty \frac{\log(1+x^7)}{(1+x^2)^2}dx
=& -\frac{7\pi}8+\frac{13\pi}8\ln2+\frac\pi2\ln\left(\cos\frac\pi{28} \cos\frac{3\pi}{28} \cos\frac{5\pi}{28}\right)\\
& -\frac\pi{28}\ln\tan\frac\pi{28}+\frac{3\pi}{28}\ln\tan\frac{3\pi}{28} -\frac{5\pi}{28}\ln\tan\frac{5\pi}{28} -\frac1{14}G\\
 \int_0^\infty \frac{\log(1+x^8)}{(1+x^2)^2}dx
=& -\pi+\pi\ln\left(\sqrt2+\sqrt{2+\sqrt2}\right)\\
\end{align}
A: 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}
A: We can attack the other integral
$$I = \int_0^{\infty} dx \frac{\log{(1+x^3)}}{(1+x^2)^2}$$
in a manner similar to what @O.L. outlined in his answer for the other case, but with a different contour.  To wit, consider
$$\oint_C dz \frac{\log{(1+z^3)} \log{z}}{(1+z^2)^2}$$
where $C$ is the following contour

This is a keyhole contour about the positive real axis, but with additional keyholes about the branch points at $z=e^{i \pi/3}$, $z=-1$, and $z=e^{i 5 \pi/3}$.  There are poles of order $2$ at $z=\pm i$.
I will outline the procedure for evaluation.  The integral about the circular arcs, large and small, go to zero as the radii go to $\infty$ and $0$, respectively.  Each of the branch points introduces a jump of $i 2 \pi$ due to the logarithm in the integrand.  By the residue theorem, we have
$$-i 2 \pi \int_0^{\infty} dx \frac{\log{(1+x^3)}}{(1+x^2)^2} - i 2 \pi \int_{e^{i \pi/3}}^{\infty e^{i \pi/3}} dt \frac{\log{t}}{(1+t^2)^2} \\ - i 2 \pi \int_{e^{i \pi}}^{\infty e^{i \pi}} dt \frac{\log{t}}{(1+t^2)^2} - i 2 \pi \int_{e^{i 5 \pi/3}}^{\infty e^{i 5 \pi/3}} dt \frac{\log{t}}{(1+t^2)^2} = \\ i 2 \pi \sum_{\pm} \frac{d}{dz} \left[\frac{\log{(1+z^3)} \log{z}}{(z\pm i)^2} \right]_{z=\pm i} $$
Without going into too much detail, I will illustrate how the integrals are done by evaluating one of them.  Consider
$$\int_{e^{i \pi}}^{\infty e^{i \pi}} dt \frac{\log{t}}{(1+t^2)^2} = -\int_1^{\infty} dy \frac{\log{y}+i \pi}{(1+y^2)^2}$$
Now,
$$\int_1^{\infty} \frac{dy}{(1+y^2)^2} = \int_{\pi/4}^{\pi/2} d\theta \cos^2{\theta} = \frac{\pi}{8}-\frac14$$
$$\begin{align}\int_1^{\infty} dy\frac{\log{y}}{(1+y^2)^2} &= -\int_0^1 du \frac{u^2 \log{u}}{(1+u^2)^2}\\ &= -\sum_{k=0}^{\infty} (-1)^k (k+1) \int_0^1 u^{2 k+2} \log{u} \\ &= \sum_{k=0}^{\infty} (-1)^k \frac{k+1}{(2 k+3)^2} \\ &= \frac{G}{2} - \frac{\pi}{8}\end{align}$$
so that
$$\int_{e^{i \pi}}^{\infty e^{i \pi}} dt \frac{\log{t}}{(1+t^2)^2} = - \left ( \frac{G}{2} - \frac{\pi}{8} \right ) - i \pi \left ( \frac{\pi}{8}-\frac14\right ) $$
Along similar lines,
$$\int_{e^{i \pi/3}}^{\infty e^{i \pi/3}} dt \frac{\log{t}}{(1+t^2)^2} =  \frac{G}{3}-\frac{\pi }{8}+\frac{1}{12} \pi  \log \left(2+\sqrt{3}\right)+i
   \left(\frac{1}{4} \log \left(2+\sqrt{3}\right)-\frac{\pi }{6}\right)$$
$$\int_{e^{i 5 \pi/3}}^{\infty e^{i 5 \pi/3}} dt \frac{\log{t}}{(1+t^2)^2} = \frac{G}{3}-\frac{\pi }{8}-\frac{5}{12} \pi  \log \left(2+\sqrt{3}\right)+i
   \left(-\frac{5 \pi }{6}+\frac{\pi ^2}{4}-\frac{1}{4} \log
   \left(2+\sqrt{3}\right)\right)$$
Combining the integrals, I get
$$\frac{G}{6} -\frac{\pi}{8}-\frac{\pi}{3} \log{(2+\sqrt{3})} + i \left [-\frac{3 \pi}{4} + \frac{\pi^2}{8}\right ] $$
The sum of the residues on the RHS is relatively simple to evaluate; I get
$$\sum_{\pm} \frac{d}{dz} \left[\frac{\log{(1+z^3)} \log{z}}{(z\pm i)^2} \right]_{z=\pm i}  = \frac{\pi}{2}-\frac{\pi}{8}\log (2)+i \left(\frac{3 \pi }{4}-\frac{\pi ^2}{8}\right)$$
The integral we seek is then the negative of the sum of the combined integrals and the sum of the residues, which gives us
$$\int_0^{\infty} dx \frac{\log{(1+x^3)}}{(1+x^2)^2} = -\frac{G}{6} - \frac{3\pi}{8} + \frac{\pi}{8} \log{2} + \frac{\pi}{3} \log{(2+\sqrt{3})} \approx 0.320555$$
which agrees with Mathematica.  Note how the imaginary parts fortuitously canceled.
It should be understood that the above technique may be applied to the other integral.  As O.L. has demonstrated, however, one may exploit symmetry and use a less computationally demanding technique for that particular case.
A: 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.
