Integration Grandmaster - P5

The substitution $u = \sqrt{1+x^2} - x$ gives $\frac{du}{dx} \; = \; \frac{x}{\sqrt{1+x^2}} - 1 \; = \; - \frac{u}{\sqrt{1 + x^2}} \; = \; -\frac{\sqrt{u}}{\sqrt{1+x^2}\sqrt{x + \sqrt{1+x^2}}}$ and so the desired integral is equal to

$\begin{aligned} A \; = \; \int_{\sqrt{2}-1}^1 \frac{\ln u}{\sqrt{u}}\,du & = \; \Big[ 2\sqrt{u} \ln u\Big]_{\sqrt{2}-1}^1 - 2\int_{\sqrt{2}-1}^1 \frac{du}{\sqrt{u}} \; = \; \Big[2\sqrt{u} \ln u - 4\sqrt{u}\Big]_{\sqrt{2}-1}^1 \\ & = \; 2\sqrt{\sqrt{2}-1}\Big[\ln(1+\sqrt{2}) + 2\Big] - 4 \end{aligned}$ which makes $\lfloor 10^4 A\rfloor = \boxed{-2912}$ .