Let's solve the integral step by step: \[ \int \frac{3x^3 - x^2 + 2x - 4}{\sqrt{x^2 - 3x + 2}} \, dx \] ## 1. Factor and Substitute First, factor the quadratic in the denominator\[ x^2 - x + 2 = (x1)(x2) \] Let\[ u = x^2 - 3x + 2 \implies du = (2x - 3)dx \] But before substituting, let's perform polynomial division since the numerator is of higher degree than the denominator. ## 2. Polynomial Division Divide \( 3x^3 - x^2 + 2x - 4 \) by \( x^2 - 3x + 2 \): #### First step: - \( 3x^3 \div x^2 = 3x \) - Multiply: \( 3x(x^2 - 3x + 2) = 3x^3 - 9x^2 + 6x \) - Subtract: \( (3x^3 - x^2 + 2x - 4) - (3x^3 - 9x^2 + 6x) = 8x^2 - 4x - 4 \) #### Second step: - \( 8x^2 \div x^2 = 8 \) - Multiply: \( 8(x^2 - 3x + 2) = 8x^2 - 24x + 16 \) - Subtract: \( (8x^2 - 4x - 4) - (8x^2 - 24x + 16) = 20x - 20 \) So: \[ \frac{3x^3 - x^2 + 2x - 4}{x^2 - 3x + 2} = 3x + 8 + \frac{20x - 20}{x^2 - 3x + 2} \] Therefore, the integral becomes: \[ \int \frac{3x^3 - x^2 + 2x - 4}{\sqrt{x^2 - 3x + 2}} dx = \int (3x + 8) \sqrt{x^2 - 3x + 2} dx + \int \frac{20x - 20}{\sqrt{x^2 - 3x + 2}} dx \] --- ## 3. Break Into Two Parts ### Part 1: \(\int (3x + 8) \sqrt{x^2 - 3x + 2} dx\) Let \( u = x^2 - 3x + 2 \), so \( du = (2x - 3) dx \): Split \( 3x + 8 \) as: \[ 3x + 8 = \frac{3}{2}(2x - 3) + \frac{25}{2} \] Check: \[ \frac{3}{2}(2x - 3) + \frac{25}{2} = 3x - \frac{9}{2} + \frac{25}{2} = 3x + 8 \] So, \[ \int (3x + 8) \sqrt{x^2 - 3x + 2} dx = \int \left[ \frac{3}{2}(2x - 3) + \frac{25}{2} \right]\sqrt{u} dx \] \[ = \frac{3}{2} \int (2x - 3)\sqrt{u} dx + \frac{25}{2} \int \sqrt{u} dx \] Recall \( du = (2x - 3)dx \), so \( (2x - 3)dx = du \). Thus: \[ \frac{3}{2} \int (2x - 3)\sqrt{u} dx = \frac{3}{2} \int \sqrt{u} du = \frac{3}{2} \cdot \frac{2}{3} u^{3/2} = u^{3/2} \] And: \[ \frac{25}{2} \int \sqrt{u} dx = \frac{25}{2} \int u^{1/2} dx \] But \( dx = \frac{du}{2x - 3} \), so this is not directly integrable in terms of \( u \) alone. We'll leave this term and move on. --- ### Part 2: \(\int \frac{20x - 20}{\sqrt{x^2 - 3x + 2}} dx\) Factor numerator: \[ 20x - 20 = 20(x - 1) \] So: \[ \int \frac{20(x - 1)}{\sqrt{x^2 - 3x + 2}} dx \] Let’s let \( u = x^2 - 3x + 2 \), \( du = (2x - 3)dx \). Express \( x - 1 \) in terms of \( u \): \[ u = x^2 - 3x + 2 \implies x^2 - 3x + 2 = u \] \[ x^2 - 3x + 2 = u \implies x^2 - 3x = u - 2 \implies x^2 = u - 2 + 3x \implies x^2 - 3x = u - 2 \] Alternatively, let's use substitution for \( x \) in \( 20(x - 1) \): Or, better yet, split the numerator as follows: \[ 20x - 20 = 10(2x - 3) + 10 \] Check: \[ 10(2x - 3) + 10 = 20x - 30 + 10 = 20x - 20 \] Thus, \[ \int \frac{20x - 20}{\sqrt{x^2 - 3x + 2}} dx = 10 \int \frac{2x - 3}{\sqrt{x^2 - 3x + 2}} dx + 10 \int \frac{1}{\sqrt{x^2 - 3x + 2}} dx \] For the first term, use \( u = x^2 - 3x + 2 \), \( du = (2x - 3)dx \): \[ 10 \int \frac{2x - 3}{\sqrt{x^2 - 3x + 2}} dx = 10 \int \frac{du}{\sqrt{u}} = 10 \int u^{-1/2} du = 20 u^{1/2} \] The second term: \[ 10 \int \frac{1}{\sqrt{x^2 - 3x + 2}} dx \] Let’s keep this term for now. --- ## 4. Collect All Integrals So far, we have: \[ \int \frac{3x^3 - x^2 + 2x - 4}{\sqrt{x^2 - 3x + 2}} dx = u^{3/2} + \frac{25}{2} \int u^{1/2} dx + 20 u^{1/2} + 10 \int \frac{1}{\sqrt{u}} dx \] Recall: - \( u = x^2 - 3x + 2 \) - \( \int u^{1/2} dx \) and \( \int \frac{1}{\sqrt{u}} dx \) need to be rewritten in terms of \( x \). --- ## 5. Finish the Remaining Integrals ### \( \int u^{1/2} dx \) Recall \( du = (2x - 3) dx \implies dx = \frac{du}{2x - 3} \): \[ \int u^{1/2} dx = \int u^{1/2} \cdot \frac{du}{2x - 3} \] But this is not elementary in terms of \( u \) alone. Instead, revert to \( x \): \[ \int \sqrt{x^2 - 3x + 2} dx \] Complete the square: \[ x^2 - 3x + 2 = (x - 3/2)^2 - (9/4 - 2) = (x - 3/2)^2 - 1/4 \] So: \[ \int \sqrt{(x - 3/2)^2 - (1/2)^2} dx \] This is a standard integral: \[ \int \sqrt{x^2 - a^2} dx = \frac{1}{2} \left( x \sqrt{x^2 - a^2} - a^2 \ln|x + \sqrt{x^2 - a^2}| \right) + C \] Let \( y = x - 3/2 \), \( a = 1/2 \): \[ \int \sqrt{(x - 3/2)^2 - (1/2)^2} dx = \frac{1}{2} \left[ (x - 3/2) \sqrt{(x - 3/2)^2 - (1/2)^2} - \frac{1}{4} \ln \left| (x - 3/2) + \sqrt{(x - 3/2)^2 - (1/2)^2} \right| \right] + C \] ### \( \int \frac{1}{\sqrt{x^2 - 3x + 2}} dx \) Similarly, the standard form is: \[ \int \frac{1}{\sqrt{x^2 - a^2}} dx = \ln|x + \sqrt{x^2 - a^2}| + C \] So, for our shifted variable: \[ \int \frac{1}{\sqrt{x^2 - 3x + 2}} dx = \ln \left| (x - 3/2) + \sqrt{(x - 3/2)^2 - (1/2)^2} \right| + C \] --- ## 6. Write the Final Answer Recall: - \( u = x^2 - 3x + 2 \) - \( u^{3/2} = (x^2 - 3x + 2)^{3/2} \) - \( u^{1/2} = (x^2 - 3x + 2)^{1/2} \) Gathering all terms: \[ \boxed{ \begin{aligned} \int \frac{3x^3 - x^2 + 2x - 4}{\sqrt{x^2 - 3x + 2}} dx = \; & (x^2 - 3x + 2)^{3/2} \\ & + \frac{25}{4} \Bigg[ (x - \frac{3}{2}) \sqrt{(x - \frac{3}{2})^2 - (\frac{1}{2})^2} \\ & \qquad - \frac{1}{2} \ln \left| (x - \frac{3}{2}) + \sqrt{(x - \frac{3}{2})^2 - (\frac{1}{2})^2} \right| \Bigg] \\ & + 20 \sqrt{x^2 - 3x + 2} \\ & + 10 \ln \left| (x - \frac{3}{2}) + \sqrt{(x - \frac{3}{2})^2 - (\frac{1}{2})^2} \right| + C \end{aligned} } \] Where \( C \) is the constant of integration. --- ## **Summary of Steps** 1. **Polynomial division** to reduce integrand degree. 2. **Break integral** into simpler terms. 3. **Substitute and integrate** each term, using standard forms for quadratic expressions under roots. 4. **Combine results** for the final answer.