<s>[INST] Evaluate the following integral :$\int\limits_0^{\infty}\frac{\log (1+x^{4})}{\sqrt{x}(1+x)}dx$ Evaluate the following integral : 

$$I=\int\limits_0^{\infty}\frac{\log (1+x^{4})}{\sqrt{x}(1+x)}dx$$

I was tried use change variable , 
If I use $x=y^2$ integral becomes : 
$$I=2\int\limits_0^{\infty}\frac{\log (1+x^{8})}{1+x^{2}}dx$$
From here I have one idea  the derivative under sing integral but I got I difficult integration : 
$$I=2\int\limits_0^{\infty}\frac{x^{8}}{(1+ax^{8})(1+x)}dx$$
I already to see you hints or solution!
 [/INST] Using the principal branch of the logarithm, we have
$$ \begin{align} I &= \int_{0}^{\infty} \frac{\ln(1+t^{4})}{\sqrt{t}(1+t)} \, \mathrm dt = 2\int_{0}^{\infty} \frac{\ln(1+x^8)}{1+x^2} \, \mathrm dx  = \int_{-\infty}^{\infty} \frac{\ln(1+x^8)}{1+x^2} \, \mathrm dx \\ &= \sum_{n=0}^{7} \int_{-\infty}^{\infty} \frac{\ln \left(1-xe^{i \pi(2n+1)/8}\right)}{1+x^2} \, \mathrm dx \\ &= \sum_{n=0}^{3} \left(\int_{-\infty}^{\infty} \frac{\ln \left(1-xe^{i \pi(2n+1)/8}\right)}{1+x^2} \, \mathrm dx + \int_{-\infty}^{\infty} \frac{\ln \left(1 \,  {\color{red}{+}} \, xe^{i \pi(2n+1)/8}\right)}{1+x^2} \, \mathrm dx \right)\\ &= \sum_{n=0}^{3} \left(\int_{-\infty}^{\infty} \frac{\ln \left(1-xe^{i \pi(2n+1)/8}\right)}{1+x^2} \, \mathrm dx+ \int_{\infty}^{-\infty} \frac{\ln \left(1 - ue^{i \pi(2n+1)/8}\right)}{1+u^2} \,  (- \mathrm du) \right) \\&= 2 \sum_{n=0}^{3} \int_{-\infty}^{\infty} \frac{\ln \left(1 - xe^{i \pi(2n+1)/8}\right)}{1+x^2} \, \mathrm dx \\ & \stackrel{(1)}= 2\sum_{n=0}^{3}    2 \pi i \operatorname{Res} \left[ \frac{\ln \left(1-ze^{i \pi(2n+1)/8}\right)}{1+z^2}, i  \right] \\ &=  2\sum_{n=0}^{3} 2 \pi i  \, \frac{\ln \left(1-ie^{i \pi(2n+1)/8}\right)}{2i} \\ &= 2\pi \sum_{n=0}^{3} \ln \left(1-ie^{i \pi(2n+1)/8}\right)  \\ &= 2 \pi \ln   \left[ \left(1-ie^{i \pi/8} \right) \left(1-ie^{3 \pi i/8} \right) \left(1+ie^{-3 \pi i/8} \right) \left(1+ie^{- i \pi/8} \right) \right]\\ &= 2 \pi \ln \left[\left(2+ 2 \sin \left(\frac{\pi}{8} \right) \right) \left(2+ 2 \sin \left(\frac{3\pi}{8} \right) \right)\right] \\ &= 2 \pi \ln \left(4 +2 \sqrt{2-\sqrt{2}} +2 \sqrt{2+\sqrt{2}} + \sqrt{2} \right) \\ &= 2 \pi \ln \left(4 + 2 \sqrt{4 +2\sqrt{2}} + \sqrt{2}\right) \\ &= 2 \pi \ln \left(\left(\sqrt{2}+\sqrt{2+\sqrt{2}} \right)^{2}\right) \\ &= 4 \pi \ln \left(\sqrt{2}+\sqrt{2+\sqrt{2}} \right) \end{align}$$

$(1)$ The branch cut for $\ln \left(1-ze^{i \pi(2n+1)/8}\right) $ is in the lower half of the complex plane.
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