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?
 A: 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.
A: Let:
$$ f(s)=\int_{0}^{1}\frac{x^s}{\sqrt{1+x}}\,dx. $$
We have, by integration by parts:
$$ f(s+1)+f(s) = \int_{0}^{1} x^s\sqrt{1+x}\,dx = \frac{1}{s+1}\left(\sqrt{2}-\frac{1}{2}\,f(s+1)\right)$$
hence:
$$ \left(2s+3\right)\, f(s+1)+(2s+2)\,f(s) = 2\sqrt{2},$$
$f(0)=2\sqrt{2}-1$ and $\lim_{s\to +\infty}f(s)=0$. We have:
$$ f(s)=\sum_{n\geq 0}\frac{(-1)^n (n-1/2)!}{\sqrt{\pi}(s+n+1)n!} $$
and to compute:
$$ \int_{0}^{1}\frac{\log^k x}{\sqrt{x(x+1)}}\,dx $$ 
boils down to computing $f^{(k)}\left(-\frac{1}{2}\right)$. That gives:
$$ 2^{k+1}\phantom{}_{k+2} F_{k+1}\left(\frac{1}{2},\frac{1}{2},\ldots,\frac{1}{2};\frac{3}{2},\frac{3}{2},\ldots,\frac{3}{2};-1\right),$$
so $k=1$ and $k=2$ are essentially the only cases we are able to manage through hypergeometric identities.
