$\int\limits^1_0 {x\sqrt{1+x^2}} \, dx \\ =\dfrac{1}{2}\int\limits^1_0 {2x\sqrt{1+x^2}} \, dx\\ =\dfrac{1}{2}\int\limits^1_0 {\sqrt{1+x^2}} \, d(x^2+1)\\ =\dfrac{1}{2}.\dfrac{2}{3}.(1+x^2)^{\tfrac{3}{2}} \Big|^1_0\\ =\dfrac{1}{3}(\sqrt{8}-1)\\ =\dfrac{2}{3}\sqrt{2}-\dfrac{1}{3}\\ =a\sqrt{2}+b\\ =>a=\dfrac{2}{3};b=-\dfrac{1}{3}\\ =>a-b=1$