$\displaystyle\int\limits^1_0 x\sqrt{3x^2+1} \, dx\\ =\dfrac{1}{6}\displaystyle\int\limits^1_0 6x\sqrt{3x^2+1} \, dx\\ =\dfrac{1}{6}\displaystyle\int\limits^1_0 \sqrt{3x^2+1} \, d(3x^2+1)\\ =\dfrac{1}{6}.\dfrac{2}{3}.(3x^2+1)^{\dfrac{3}{2}} \Big|^1_0\\ =\dfrac{1}{9}\sqrt{(3x^2+1)^3}\Big|^1_0\\ =\dfrac{7}{9}$