1. **State the problem:** Evaluate the first integral $$\int_0^2 \frac{1}{(2x - 1)^{3/2}} \, dx$$ since the other integrals are either improper or complex and the user asked to solve the first problem completely. 2. **Recall the formula:** For integrals of the form $$\int \frac{1}{(ax + b)^n} \, dx$$, use substitution and power rule. 3. **Substitution:** Let $$u = 2x - 1$$, then $$du = 2 dx$$ or $$dx = \frac{du}{2}$$. 4. **Change limits:** When $$x=0$$, $$u=2(0)-1=-1$$; when $$x=2$$, $$u=2(2)-1=3$$. 5. **Rewrite integral:** $$\int_0^2 \frac{1}{(2x - 1)^{3/2}} \, dx = \int_{-1}^3 \frac{1}{u^{3/2}} \cdot \frac{du}{2} = \frac{1}{2} \int_{-1}^3 u^{-3/2} \, du$$ 6. **Evaluate integral:** $$\int u^{-3/2} \, du = \int u^{-\frac{3}{2}} \, du = \frac{u^{-\frac{3}{2}+1}}{-\frac{3}{2}+1} + C = \frac{u^{-\frac{1}{2}}}{-\frac{1}{2}} + C = -2 u^{-1/2} + C$$ 7. **Apply limits:** $$\frac{1}{2} \times [-2 u^{-1/2}]_{-1}^3 = \frac{1}{2} \times [-2 (3)^{-1/2} + 2 (-1)^{-1/2}] = \frac{1}{2} \times [-2/\sqrt{3} + 2/\sqrt{-1}]$$ 8. **Note:** $$\sqrt{-1} = i$$ (imaginary unit), so the integral is improper over the interval including negative values inside the root. 9. **Conclusion:** The integral $$\int_0^2 \frac{1}{(2x - 1)^{3/2}} \, dx$$ is improper because the integrand is not real-valued for $$x < 0.5$$ (since $$2x-1$$ must be positive to have real values for the power $$3/2$$). The integral diverges at $$x=0.5$$. **Final answer:** The integral $$\int_0^2 \frac{1}{(2x - 1)^{3/2}} \, dx$$ is divergent due to the singularity at $$x=0.5$$. --- **Summary:** The first integral diverges and cannot be evaluated as a finite real number.