91Ó°ÊÓ

First, we need to factor the denominator, which is a quadratic polynomial $$x^2 - 3x + 2$$. We are looking for two binomials whose product is equal to the given polynomial. We can find the factors by examining the roots of the quadratic equation and factoring: $$x^2 - 3x + 2 = (x - 1)(x - 2)$$

Now, we rewrite the integrand using partial fractions. The integrand will have the form $$\frac{3x^2+4x-6}{(x-1)(x-2)} = \frac{A}{x-1}+\frac{B}{x-2}$$. To find A and B, we need to eliminate the denominators by multiplying the entire equation by $$(x-1)(x-2)$$, which results in: $$3x^2 + 4x - 6 = A(x-2) + B(x-1)$$ Now, we need to find values for A and B. We can do this by substituting suitable values of x and solving the resulting system of equations. Let's substitute x = 1: $$3(1)^2 + 4(1) - 6 = A(1 - 2) + B(1 - 1)$$ This simplifies to 1 = -A which means A = -1. Now, let's substitute x = 2: $$3(2)^2 + 4(2) - 6 = A(2 - 1) + B(2 - 2)$$ This simplifies to 10 = -B which means B = 10. So we can write our integrand as: $$\frac{-1}{x-1}+\frac{10}{x-2}$$

Now that we have rewritten the integrand, we can integrate each term separately: $$\int \frac{-1}{x-1} dx + \int \frac{10}{x-2} dx$$ Using the properties of integrals and integrating both terms separately, we get: $$-1 \int \frac{1}{x-1} dx + 10 \int \frac{1}{x-2} dx$$ These integrals are both of the form $$\int \frac{1}{x-a} dx$$, which integrate to $$\ln |x-a| + C$$. As a result, we have: $$-1(\ln |x-1| + C_1) + 10(\ln |x-2| + C_2)$$

Let's simplify the result and combine the constants of integration (\({C_1}\) and \({C_2}\)) into a single constant (C): $$\ln |x-1|^{-1} + \ln |x-2|^{10} + C$$ Our final solution for the integral$$\int \frac{3 x^{2}+4 x-6}{x^{2}-3 x+2} dx$$ is: $$\ln |x-1|^{-1} + \ln |x-2|^{10} + C$$