1. **State the problem:** We need to evaluate the integral $$\int \frac{2x^3 - 4x + 6}{2x^2 - 3} \, dx.$$\n\n2. **Analyze the integrand:** The numerator is a cubic polynomial and the denominator is quadratic. Since the degree of the numerator (3) is greater than the degree of the denominator (2), we perform polynomial long division first.\n\n3. **Polynomial division:** Divide $2x^3 - 4x + 6$ by $2x^2 - 3$.\n\n- Leading term division: $\frac{2x^3}{2x^2} = x$.\n- Multiply divisor by $x$: $x(2x^2 - 3) = 2x^3 - 3x$.\n- Subtract: $(2x^3 - 4x + 6) - (2x^3 - 3x) = -x + 6$.\n\nSo, $$\frac{2x^3 - 4x + 6}{2x^2 - 3} = x + \frac{-x + 6}{2x^2 - 3}.$$\n\n4. **Rewrite the integral:** $$\int \frac{2x^3 - 4x + 6}{2x^2 - 3} \, dx = \int x \, dx + \int \frac{-x + 6}{2x^2 - 3} \, dx.$$\n\n5. **Integrate the first term:** $$\int x \, dx = \frac{x^2}{2} + C.$$\n\n6. **Focus on the second integral:** $$\int \frac{-x + 6}{2x^2 - 3} \, dx.$$\n\n7. **Use substitution:** Let $$u = 2x^2 - 3,$$ then $$du = 4x \, dx \implies x \, dx = \frac{du}{4}.$$\n\nRewrite numerator: $$-x + 6 = -x + 6.$$ We want to express the integral in terms of $u$ and $du$. Split the integral:\n$$\int \frac{-x}{2x^2 - 3} \, dx + \int \frac{6}{2x^2 - 3} \, dx.$$\n\n8. **First part:** $$\int \frac{-x}{2x^2 - 3} \, dx = -\int \frac{x}{u} \, dx.$$ Using substitution, $$x \, dx = \frac{du}{4},$$ so\n$$-\int \frac{x}{u} \, dx = -\int \frac{1}{u} \cdot \frac{du}{4} = -\frac{1}{4} \int \frac{1}{u} \, du = -\frac{1}{4} \ln|u| + C = -\frac{1}{4} \ln|2x^2 - 3| + C.$$\n\n9. **Second part:** $$\int \frac{6}{2x^2 - 3} \, dx = 6 \int \frac{1}{2x^2 - 3} \, dx.$$\n\nRewrite denominator: $$2x^2 - 3 = 2\left(x^2 - \frac{3}{2}\right).$$\n\n10. **Integral of the form:** $$\int \frac{1}{x^2 - a^2} \, dx = \frac{1}{2a} \ln \left| \frac{x - a}{x + a} \right| + C,$$ for $a^2 > 0$. Here, $a = \sqrt{\frac{3}{2}} = \frac{\sqrt{6}}{2}$.\n\n11. **Apply formula:**\n$$6 \int \frac{1}{2(x^2 - \frac{3}{2})} \, dx = 3 \int \frac{1}{x^2 - (\frac{\sqrt{6}}{2})^2} \, dx = 3 \cdot \frac{1}{2a} \ln \left| \frac{x - a}{x + a} \right| + C = \frac{3}{2a} \ln \left| \frac{x - a}{x + a} \right| + C.$$\n\nSubstitute $a = \frac{\sqrt{6}}{2}$:\n$$\frac{3}{2 \cdot \frac{\sqrt{6}}{2}} = \frac{3}{\sqrt{6}} = \frac{3 \sqrt{6}}{6} = \frac{\sqrt{6}}{2}.$$\n\nSo,\n$$6 \int \frac{1}{2x^2 - 3} \, dx = \frac{\sqrt{6}}{2} \ln \left| \frac{x - \frac{\sqrt{6}}{2}}{x + \frac{\sqrt{6}}{2}} \right| + C.$$\n\n12. **Combine all parts:**\n$$\int \frac{2x^3 - 4x + 6}{2x^2 - 3} \, dx = \frac{x^2}{2} - \frac{1}{4} \ln|2x^2 - 3| + \frac{\sqrt{6}}{2} \ln \left| \frac{x - \frac{\sqrt{6}}{2}}{x + \frac{\sqrt{6}}{2}} \right| + C.$$