<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] In my post, I had found a formula for the integral
$$I(a)=\int_{0}^{\infty} \frac{\ln \left(x^{4}+2 x^{2} \cos 2 a+1\right)}{1+x^{2}} d x=2 \pi \ln \left(2 \cos \frac{a}{2}\right)$$
Factorise $x^8+1= \left(x^4+\sqrt{2} x^2+1\right)\left(x^4-\sqrt{2} x^2+1\right)$ split our integral into two integrals with quartic polynomials.
$$
\begin{aligned}
I&=2 \int_0^{\infty} \frac{\ln \left[\left(x^4+\sqrt{2} x^2+1\right)\left(x^4-\sqrt{2} x^2+1\right)\right]}{1+x^2} d x \\
&=2\left[\int_0^{\infty} \frac{\ln \left(x^4+\sqrt{2} x^2+1\right)}{1+x^2} d x+\int_0^{\infty} \frac{\ln \left(x^4-\sqrt{2} x^2+1\right)}{1+x^2} d x\right] \\
&=4 \pi\left\{\ln \left[2 \cos \left(\frac{1}{4} \cos ^{-1}\left(\frac{\sqrt{2}}{2}\right)\right)\right]+\ln \left[2 \cos \left(\frac{1}{4} \cos ^{-1}\left(\frac{-\sqrt{2}}{2}\right)\right)\right]\right\} \\
&=4 \pi\left\{\ln \left(2 \cos \frac{\pi}{16}\right)+\ln \left(2 \cos \frac{3 \pi}{16}\right)\right\} \\
&=4 \pi \ln (\sqrt{2}+\sqrt{2+\sqrt{2}})
\end{aligned}
$$
</s>