$\begin{align}  & \int_{0}^{1}{\frac{du}{\sqrt{3-2u}}} \\ & \text{Let }x=3-2u \\ & \frac{dx}{du}=-2,\text{ }du=-\frac{dx}{2} \\ & \int{\frac{du}{\sqrt{3-2u}}}=\int{\frac{dx}{2\sqrt{x}}}=\frac{1}{2}\int{{{x}^{-\tfrac{1}{2}}}dx} \\ & \int{\frac{du}{\sqrt{3-2u}}}=\frac{1}{2}\left[ \frac{{{x}^{-\tfrac{1}{2}+1}}}{-\tfrac{1}{2}+1} \right]=\frac{1}{2}\left[ \frac{{{x}^{\tfrac{1}{2}}}}{\tfrac{1}{2}} \right]={{x}^{\tfrac{1}{2}}}+C \\ & \int{\frac{du}{\sqrt{3-2u}}}={{x}^{\tfrac{1}{2}}}=\sqrt{3-2u} \\ & \int_{0}^{1}{\frac{du}{\sqrt{3-2u}}}=\left[ \sqrt{3-2u} \right]_{0}^{1} \\ & \int_{0}^{1}{\frac{du}{\sqrt{3-2u}}}=\left[ \sqrt{3-2(1)}-\sqrt{3-2(0)} \right] \\ & \int_{0}^{1}{\frac{du}{\sqrt{3-2u}}}=1-\sqrt{3} \\\end{align}$