$\begin{align}  & \int_{0}^{\sqrt{\tfrac{1}{3}}}{\frac{dx}{(2-3{{x}^{2}})}}=\int_{0}^{\sqrt{\tfrac{1}{3}}}{\frac{dx}{{{(\sqrt{2})}^{2}}-{{(\sqrt{3}x)}^{2}}}} \\ & \int_{0}^{\sqrt{\tfrac{1}{3}}}{\frac{dx}{(2-3{{x}^{2}})}}=\int_{0}^{\sqrt{\tfrac{1}{3}}}{\frac{dx}{(\sqrt{2})-(\sqrt{3x})(\sqrt{2}-\sqrt{3}x)}} \\ & \text{Resolving }\frac{1}{(\sqrt{2})-(\sqrt{3x})(\sqrt{2}-\sqrt{3}x)}\text{ into partial fraction} \\ & \frac{dx}{(\sqrt{2})-(\sqrt{3x})(\sqrt{2}-\sqrt{3}x)}=\frac{A}{\sqrt{2}-\sqrt{3}x}+\frac{B}{\sqrt{2}+\sqrt{3}x} \\ & 1=A(\sqrt{2}+\sqrt{3}x)+B(\sqrt{2}-\sqrt{3}x) \\ & \text{Set }x=\frac{\sqrt{2}}{\sqrt{3}} \\ & A=\frac{1}{2\sqrt{2}}=\frac{\sqrt{2}}{4} \\ & \text{Set }x=-\frac{\sqrt{2}}{\sqrt{3}} \\ & B=\frac{1}{2\sqrt{2}}=\frac{\sqrt{2}}{4} \\ & \int_{0}^{\sqrt{\tfrac{1}{3}}}{\frac{dx}{(2-3{{x}^{2}})}}=\int_{0}^{\sqrt{\tfrac{1}{3}}}{\left( \frac{\sqrt{2}}{4(\sqrt{2}-\sqrt{3}x)}+\frac{\sqrt{2}}{4(\sqrt{2}+\sqrt{3}x)} \right)dx} \\ & \int_{0}^{\sqrt{\tfrac{1}{3}}}{\frac{dx}{(2-3{{x}^{2}})}}=\frac{\sqrt{2}}{4}\int_{0}^{\sqrt{\tfrac{1}{3}}}{\left( \frac{1}{4\sqrt{2}-\sqrt{3}x}+\frac{1}{4\sqrt{2}+\sqrt{3}x} \right)dx} \\ & \int_{0}^{\sqrt{\tfrac{1}{3}}}{\frac{dx}{(2-3{{x}^{2}})}}=\frac{\sqrt{2}}{4}\left( -\frac{1}{\sqrt{3}}\ln (\sqrt{2}-\sqrt{3}x)+\frac{1}{\sqrt{3}}\ln (\sqrt{20}+\sqrt{3}x) \right) \\ & =\frac{\sqrt{2}}{4\sqrt{3}}\left( \ln \frac{\sqrt{2}+\sqrt{3}x}{\sqrt{2}-\sqrt{3}x} \right)_{0}^{\sqrt{\tfrac{1}{3}}}=\frac{\sqrt{2}}{4\sqrt{3}}\left[ \ln \frac{\sqrt{2}+1}{\sqrt{2}-1}-\ln 0 \right] \\ & =\frac{\sqrt{2}}{4\sqrt{3}}\left( \ln \frac{\sqrt{2}+1}{\sqrt{2}-1} \right)=\frac{\sqrt{6}}{12}\left[ \ln {{(\sqrt{2}+1)}^{2}} \right]+C \\\end{align}$