1. Stating the problem: Evaluate the integral $$\int \frac{3\sqrt{x} - 1}{5x - 6} \, dx.$$\n\n2. Rewrite the integral by expressing \(\sqrt{x}\) as \(x^{1/2}\):\n$$\int \frac{3x^{1/2} - 1}{5x - 6} \, dx.$$\n\n3. Since the integrand is a rational function involving fractional powers, consider substitution or decomposing numerator to simplify.\n\n4. Attempt substitution: Let \(u = 5x - 6\), then \(du = 5 \, dx \Rightarrow dx = \frac{du}{5}\). Express \(x\) and \(x^{1/2}\) in terms of \(u\):\n$$x = \frac{u + 6}{5}, \quad x^{1/2} = \sqrt{\frac{u + 6}{5}} = \frac{\sqrt{u + 6}}{\sqrt{5}}.$$\n\n5. Substitute into integral:\n$$\int \frac{3 \frac{\sqrt{u + 6}}{\sqrt{5}} - 1}{u} \cdot \frac{du}{5} = \frac{1}{5} \int \frac{3 \sqrt{u + 6} / \sqrt{5} - 1}{u} du = \frac{1}{5} \int \left( \frac{3 \sqrt{u + 6}}{u \sqrt{5}} - \frac{1}{u} \right) du.$$\n\n6. Simplify constants:\n$$= \frac{1}{5} \int \left( \frac{3}{\sqrt{5}} \cdot \frac{\sqrt{u + 6}}{u} - \frac{1}{u} \right) du = \frac{3}{5 \sqrt{5}} \int \frac{\sqrt{u + 6}}{u} du - \frac{1}{5} \int \frac{1}{u} du.$$\n\n7. The integral \(\int \frac{1}{u} du = \ln |u| + C\.\) For the first integral, consider the substitution \(t = \sqrt{u+6}\), so \(u = t^{2} - 6\), \(du = 2t dt\). Then substitute:\n$$\int \frac{\sqrt{u + 6}}{u} du = \int \frac{t}{t^{2} - 6} \cdot 2t dt = 2 \int \frac{t^{2}}{t^{2} - 6} dt.$$\n\n8. Simplify integrand:\n$$2 \int \frac{t^{2}}{t^{2} - 6} dt = 2 \int \left(1 + \frac{6}{t^{2} - 6} \right) dt = 2 \int dt + 12 \int \frac{1}{t^{2} - 6} dt = 2t + 12 \int \frac{1}{t^{2} - 6} dt.$$\n\n9. For the integral \(\int \frac{1}{t^{2} - 6} dt\), write denominator as \((t - \sqrt{6})(t + \sqrt{6})\). Use partial fraction decomposition: \(\frac{1}{t^{2} - 6} = \frac{A}{t - \sqrt{6}} + \frac{B}{t + \sqrt{6}}\).\n\n10. Solve for \(A\) and \(B\): multiply both sides by \(t^{2} - 6\):\n$$1 = A(t + \sqrt{6}) + B(t - \sqrt{6}).$$\nSet \(t = \sqrt{6}\):\n$$1 = A(2 \sqrt{6}) + B (0) \Rightarrow A = \frac{1}{2 \sqrt{6}}.$$\nSet \(t = - \sqrt{6}\):\n$$1 = A(0) + B (-2 \sqrt{6}) \Rightarrow B = -\frac{1}{2 \sqrt{6}}.$$\n\n11. Therefore,\n$$\int \frac{1}{t^{2} - 6} dt = \frac{1}{2 \sqrt{6}} \int \frac{1}{t - \sqrt{6}} dt - \frac{1}{2 \sqrt{6}} \int \frac{1}{t + \sqrt{6}} dt = \frac{1}{2 \sqrt{6}} \ln |t - \sqrt{6}| - \frac{1}{2 \sqrt{6}} \ln |t + \sqrt{6}| + C.$$\n\n12. Combine logs:\n$$= \frac{1}{2 \sqrt{6}} \ln \left| \frac{t - \sqrt{6}}{t + \sqrt{6}} \right| + C.$$\n\n13. Substitute back to \(t = \sqrt{u + 6}\):\n$$2t + 12 \int \frac{1}{t^{2} - 6} dt = 2 \sqrt{u + 6} + 12 \cdot \frac{1}{2 \sqrt{6}} \ln \left| \frac{\sqrt{u + 6} - \sqrt{6}}{\sqrt{u + 6} + \sqrt{6}} \right| + C = 2 \sqrt{u + 6} + \frac{6}{\sqrt{6}} \ln \left| \frac{\sqrt{u + 6} - \sqrt{6}}{\sqrt{u + 6} + \sqrt{6}} \right| + C.$$\n\n14. Simplify \(\frac{6}{\sqrt{6}} = \sqrt{6} \cdot 6 / 6 = \sqrt{6} \cdot 1 = \sqrt{6}\):\n$$= 2 \sqrt{u + 6} + \sqrt{6} \ln \left| \frac{\sqrt{u + 6} - \sqrt{6}}{\sqrt{u + 6} + \sqrt{6}} \right| + C.$$\n\n15. Recall step 6, multiply back to get the original integral:\n$$\frac{3}{5 \sqrt{5}} \cdot (2 \sqrt{u + 6} + \sqrt{6} \ln \left| \frac{\sqrt{u + 6} - \sqrt{6}}{\sqrt{u + 6} + \sqrt{6}} \right|) - \frac{1}{5} \ln |u| + C.$$\n\n16. Substitute \(u = 5x - 6\) back:\n$$= \frac{3}{5 \sqrt{5}} \cdot 2 \sqrt{5x} + \frac{3}{5 \sqrt{5}} \cdot \sqrt{6} \ln \left| \frac{\sqrt{5x} - \sqrt{6}}{\sqrt{5x} + \sqrt{6}} \right| - \frac{1}{5} \ln |5x - 6| + C = \frac{6}{5 \sqrt{5}} \sqrt{5x} + \frac{3 \sqrt{6}}{5 \sqrt{5}} \ln \left| \frac{\sqrt{5x} - \sqrt{6}}{\sqrt{5x} + \sqrt{6}} \right| - \frac{1}{5} \ln |5x - 6| + C.$$\n\n17. Simplify \(\frac{6}{5 \sqrt{5}} \sqrt{5x} = \frac{6}{5} \sqrt{x}\) as \(\frac{6}{5 \sqrt{5}} \cdot \sqrt{5} \sqrt{x} = \frac{6}{5} \sqrt{x} \).\n\nFinal answer:\n$$\int \frac{3\sqrt{x} - 1}{5x - 6} dx = \frac{6}{5} \sqrt{x} + \frac{3 \sqrt{6}}{5 \sqrt{5}} \ln \left| \frac{\sqrt{5x} - \sqrt{6}}{\sqrt{5x} + \sqrt{6}} \right| - \frac{1}{5} \ln |5x - 6| + C.$$