Not the toughest integral, not the easiest one Perhaps it's not amongst the toughest integrals, but it's interesting  to try to find an elegant
approach for the integral 
$$I_1=\int_0^1 \frac{\log (x)}{\sqrt{x (x+1)}} \, dx$$
$$=4 \text{Li}_2\left(-\sqrt{2}\right)-4 \text{Li}_2\left(-1-\sqrt{2}\right)+2 \log ^2\left(1+\sqrt{2}\right)-4 \log \left(2+\sqrt{2}\right) \log \left(1+\sqrt{2}\right)-\frac{\pi ^2}{3}$$
Do you see any such a way? Then I wonder if we can think of some elegant ways for the evaluation of the quadratic and cubic versions, that is 
$$I_2=\int_0^1 \frac{\log^2 (x)}{\sqrt{x (x+1)}} \, dx$$
$$I_3=\int_0^1 \frac{\log^3 (x)}{\sqrt{x (x+1)}} \, dx.$$
How far can we possibly go with the generalization such that we can get integrals in closed form?

By making the Euler substitution $\sqrt{x^{2}+x} = x+t$, we find
$$ \begin{align} \int_{0}^{1} \frac{\log(x)}{\sqrt{x^{2}+x}} \, dx &= 2 \int_{0}^{\sqrt{2}-1} \frac{\log \left(\frac{t^{2}}{1-2t}\right)}{1-2t} \, dt \\ &= 4 \int_{0}^{\sqrt{2}-1} \frac{\log t}{1-2t} \, dt - 2\int_{0}^{\sqrt{2}-1} \frac{\log(1-2t)}{1-2t} \, dt. \end{align}$$
The first integral can be evaluated by integrating by parts.
$$\begin{align}  \int_{0}^{\sqrt{2}-1} \frac{\log t}{1-2t} \, dt &= -\frac{1}{2}\log(t) \log(1-2t) \Bigg|^{\sqrt{2}-1}_{0} + \frac{1}{2}\int_{0}^{\sqrt{2}-1} \frac{\log(1-2t)}{t} \, dt \\ &= -\frac{1}{2}\log(\sqrt{2}-1) \log(3 - 2 \sqrt{2}) - \frac{1}{2}\text{Li}_{2} \big(2(\sqrt{2}-1)\big) \end{align}$$
Therefore,
$$\begin{align}\int_{0}^{1} \frac{\log(x)}{\sqrt{x^{2}+x}} \, dx &=  -2 \log(\sqrt{2}-1) \log(3 - 2 \sqrt{2}) - 2 \text{Li}_{2} \big( 2(\sqrt{2}-1) \big) + \frac{1}{2} \log^{2}(3- 2 \sqrt{2}) \\ &\approx -3.8208072259. \end{align}$$
EDIT:
David H used an Euler substitution to evaluate a similar but much more difficult integral here.
