1. **State the problem:** We need to find the integral of the function $f(x) = (1 - x) \sqrt{2x - 3}$. That is, compute $$\int (1 - x) \sqrt{2x - 3} \, dx.$$\n\n2. **Rewrite the integral:** Let us write the integral as $$\int (1 - x)(2x - 3)^{\frac{1}{2}} \, dx.$$\n\n3. **Use substitution:** Set $$u = 2x - 3,$$ then $$du = 2 \, dx \implies dx = \frac{du}{2}.$$ Also, express $x$ in terms of $u$: $$x = \frac{u + 3}{2}.$$\n\n4. **Rewrite the integrand in terms of $u$:**\n$$1 - x = 1 - \frac{u + 3}{2} = \frac{2 - (u + 3)}{2} = \frac{-u - 1}{2} = -\frac{u + 1}{2}.$$\n\n5. **Substitute into the integral:**\n$$\int (1 - x)(2x - 3)^{\frac{1}{2}} \, dx = \int -\frac{u + 1}{2} u^{\frac{1}{2}} \cdot \frac{du}{2} = -\frac{1}{4} \int (u + 1) u^{\frac{1}{2}} \, du.$$\n\n6. **Simplify the integrand:**\n$$ (u + 1) u^{\frac{1}{2}} = u^{\frac{3}{2}} + u^{\frac{1}{2}}.$$\n\n7. **Integral becomes:**\n$$ -\frac{1}{4} \int \left(u^{\frac{3}{2}} + u^{\frac{1}{2}}\right) du = -\frac{1}{4} \left( \int u^{\frac{3}{2}} du + \int u^{\frac{1}{2}} du \right).$$\n\n8. **Integrate each term:**\n$$\int u^{\frac{3}{2}} du = \frac{u^{\frac{5}{2}}}{\frac{5}{2}} = \frac{2}{5} u^{\frac{5}{2}},$$\n$$\int u^{\frac{1}{2}} du = \frac{u^{\frac{3}{2}}}{\frac{3}{2}} = \frac{2}{3} u^{\frac{3}{2}}.$$\n\n9. **Substitute back:**\n$$ -\frac{1}{4} \left( \frac{2}{5} u^{\frac{5}{2}} + \frac{2}{3} u^{\frac{3}{2}} \right) + C = -\frac{1}{4} \left( \frac{2}{5} u^{\frac{5}{2}} + \frac{2}{3} u^{\frac{3}{2}} \right) + C.$$\n\n10. **Simplify constants:**\n$$ = -\left( \frac{1}{10} u^{\frac{5}{2}} + \frac{1}{6} u^{\frac{3}{2}} \right) + C = -\frac{1}{10} u^{\frac{5}{2}} - \frac{1}{6} u^{\frac{3}{2}} + C.$$\n\n11. **Rewrite in terms of $x$:**\n$$u = 2x - 3,$$ so\n$$\boxed{\int (1 - x) \sqrt{2x - 3} \, dx = -\frac{1}{10} (2x - 3)^{\frac{5}{2}} - \frac{1}{6} (2x - 3)^{\frac{3}{2}} + C}.$$