To evaluate \(\int \frac{3x^3 - x^2 + 2x - 4}{\sqrt{x^2 - 3x + 2}} \, dx\): 1. **Factor the denominator**: \[ x^2 - 3x + 2 = (x - 1)(x - 2) \] 2. **Use substitution for the square root**: Let \(t = \sqrt{x^2 - 3x + 2}\). Alternatively, since the denominator involves a quadratic, consider substitution to simplify the expression. 3. **Complete the square inside the square root**: \[ x^2 - 3x + 2 = (x - \frac{3}{2})^2 - \frac{9}{4} + 2 = (x - \frac{3}{2})^2 - \frac{1}{4} \] 4. **Substitute \(u = x - \frac{3}{2}\)**: \[ du = dx \] The integral becomes: \[ \int \frac{3x^3 - x^2 + 2x - 4}{\sqrt{u^2 - \frac{1}{4}}} \, dx \] But since \(x = u + \frac{3}{2}\), rewrite numerator in terms of \(u\). 5. **Rewrite numerator in terms of \(u\)**: - \(x = u + \frac{3}{2}\) - \(x^2 = u^2 + 3u + \frac{9}{4}\) - \(x^3 = u^3 + \frac{9}{2} u^2 + \frac{27}{4} u + \frac{27}{8}\) Plug into numerator: \[ 3x^3 - x^2 + 2x - 4 \] becomes a polynomial in \(u\). 6. **Perform polynomial expansion**: - \(3x^3 = 3u^3 + \frac{27}{2} u^2 + \frac{81}{4} u + \frac{81}{8}\) - \(-x^2 = -u^2 - 3u - \frac{9}{4}\) - \(2x = 2u + 3\) - \(-4\) stays the same Sum: \[ (3u^3) + \left(\frac{27}{2} u^2 - u^2\right) + \left(\frac{81}{4} u - 3u + 2u\right) + \left(\frac{81}{8} - \frac{9}{4} + 3 - 4\right) \] Simplify each group: - \(3u^3\) - \(\frac{27}{2} u^2 - u^2 = \frac{27}{2} u^2 - \frac{2}{2} u^2 = \frac{25}{2} u^2\) - \(\frac{81}{4} u - 3u + 2u = \frac{81}{4} u - \frac{12}{4} u + \frac{8}{4} u = \frac{77}{4} u\) - Constants: \(\frac{81}{8} - \frac{18}{8} + \frac{24}{8} - \frac{32}{8} = \frac{81 - 18 + 24 - 32}{8} = \frac{55}{8}\) So numerator: \[ 3u^3 + \frac{25}{2} u^2 + \frac{77}{4} u + \frac{55}{8} \] 7. **Rewrite the integral**: \[ \int \frac{3u^3 + \frac{25}{2} u^2 + \frac{77}{4} u + \frac{55}{8}}{\sqrt{u^2 - \frac{1}{4}}} du \] 8. **Express as sum of integrals**: \[ \int \frac{3u^3}{\sqrt{u^2 - \frac{1}{4}}} du + \frac{25}{2} \int \frac{u^2}{\sqrt{u^2 - \frac{1}{4}}} du + \frac