1. **State the problem:** We want to evaluate the integral $$\int \frac{x^3 + 2x}{x^2 - 5x + 6} \, dx$$ using partial fractions. 2. **Factor the denominator:** The denominator factors as $$x^2 - 5x + 6 = (x-2)(x-3)$$. 3. **Simplify the integrand:** Since the degree of the numerator (3) is greater than the degree of the denominator (2), perform polynomial division first. Divide $$x^3 + 2x$$ by $$x^2 - 5x + 6$$: - Leading term division: $$\frac{x^3}{x^2} = x$$ - Multiply divisor by $$x$$: $$x(x^2 - 5x + 6) = x^3 - 5x^2 + 6x$$ - Subtract: $$(x^3 + 2x) - (x^3 - 5x^2 + 6x) = 5x^2 - 4x$$ Now divide $$5x^2 - 4x$$ by $$x^2 - 5x + 6$$: - Leading term division: $$\frac{5x^2}{x^2} = 5$$ - Multiply divisor by 5: $$5(x^2 - 5x + 6) = 5x^2 - 25x + 30$$ - Subtract: $$(5x^2 - 4x) - (5x^2 - 25x + 30) = 21x - 30$$ So the division gives: $$\frac{x^3 + 2x}{x^2 - 5x + 6} = x + 5 + \frac{21x - 30}{(x-2)(x-3)}$$ 4. **Set up partial fractions for the remainder:** $$\frac{21x - 30}{(x-2)(x-3)} = \frac{A}{x-2} + \frac{B}{x-3}$$ Multiply both sides by $$(x-2)(x-3)$$: $$21x - 30 = A(x-3) + B(x-2)$$ 5. **Find A and B:** - Let $$x=3$$: $$21(3) - 30 = A(0) + B(1) \Rightarrow 63 - 30 = B \Rightarrow B = 33$$ - Let $$x=2$$: $$21(2) - 30 = A(-1) + B(0) \Rightarrow 42 - 30 = -A \Rightarrow 12 = -A \Rightarrow A = -12$$ 6. **Rewrite the integral:** $$\int \frac{x^3 + 2x}{x^2 - 5x + 6} \, dx = \int (x + 5) \, dx + \int \frac{-12}{x-2} \, dx + \int \frac{33}{x-3} \, dx$$ 7. **Integrate each term:** - $$\int (x + 5) \, dx = \frac{x^2}{2} + 5x + C_1$$ - $$\int \frac{-12}{x-2} \, dx = -12 \ln|x-2| + C_2$$ - $$\int \frac{33}{x-3} \, dx = 33 \ln|x-3| + C_3$$ 8. **Combine constants and write final answer:** $$\int \frac{x^3 + 2x}{x^2 - 5x + 6} \, dx = \frac{x^2}{2} + 5x - 12 \ln|x-2| + 33 \ln|x-3| + C$$