<s>[INST] Evaluate an absolute monster integral $\int\limits_{0}^{1} \frac{\log(1-x+x^2)}{\sqrt{x}(1+x)}\mathrm{d}x.$ I want to figure out a way to evaluate $$\int\limits_{0}^{1} \frac{\log(1-x+x^2)}{\sqrt{x}(1+x)}\mathrm{d}x.$$
I tried to substitute $x = u^2$ and cancel the square root in the denominator, getting $$2\int\limits_{0}^{1} \frac{\log(1-u^2+u^4)}{1+u^2}\mathrm{d}u.$$But now I am stuck again.
 [/INST] Note
$$I=\int\limits_{0}^{1} \frac{\ln(1-x+x^2)}{\sqrt{x}(1+x)}{d}x
\overset{t=\sqrt x} = 2\int\limits_{0}^{1} \frac{\ln(1-t^2+t^4)}{1+t^2}dt
$$
Let $J(a) = \int_0^1 \frac{\ln(\frac14(1+t^2)^2 \sec^2a -t^2)}{1+t^2}dt
$.
Then, $J(0) = \int_0^1 \frac{2\ln\frac{1-t^2}2}{1+t^2}dt=
-2G$
$$J’(a)= \int_0^1 \left(\frac{\tan a}{t^2+2t\cos a+1} + \frac{\tan a}{t^2-2t\cos a+1} \right)dt=\frac\pi2 \sec a
$$
$$J(\frac\pi6) = \int\limits_{0}^{1} \frac{\ln(1-t^2+t^4)-\ln3}{1+t^2}dt=\frac12I-\frac\pi4\ln3
$$
Thus
\begin{align}
I= \frac\pi2\ln3+2J(\frac\pi6)
= \frac\pi2\ln3+2J(0) +2\int_0^{\pi/6} J’(a)da
=\pi \ln3-4G
\end{align}
</s>