Evaluate the integral? : # int_3^4 (x^2-3x+6)/(x(x-2)(x-1)) dx#

We first need to find the partial fraction decomposition of the integrand, which will be of the form;

And so:

Which we can now integrate as we know that

Hence,

To evaluate the integral (\int_{3}^{4} \frac{x^2 - 3x + 6}{x(x - 2)(x - 1)} , dx), we can first perform partial fraction decomposition. After decomposing the fraction into simpler fractions, we can integrate each term separately. Then, we evaluate the integral over the given bounds.

The partial fraction decomposition of the given expression yields:

(\frac{x^2 - 3x + 6}{x(x - 2)(x - 1)} = \frac{A}{x} + \frac{B}{x - 2} + \frac{C}{x - 1})

Solving for (A), (B), and (C), we find (A = 2), (B = -1), and (C = 1).

So, the integral becomes:

(\int_{3}^{4} \left(\frac{2}{x} - \frac{1}{x - 2} + \frac{1}{x - 1}\right) , dx)

Integrating each term separately, we get:

(2\ln|x| - \ln|x - 2| + \ln|x - 1|)

Now, we evaluate this expression at the upper and lower bounds:

At (x = 4), we have:

(2\ln|4| - \ln|4 - 2| + \ln|4 - 1| = 2\ln(4) - \ln(2) + \ln(3))

At (x = 3), we have:

(2\ln|3| - \ln|3 - 2| + \ln|3 - 1| = 2\ln(3) - \ln(1) + \ln(2))

Therefore, the value of the integral is (2\ln(4) - \ln(2) + \ln(3) - 2\ln(3) + \ln(1) - \ln(2)). Simplifying, we get:

(2\ln\left(\frac{4 \cdot 3}{3^2}\right) - \ln(2) + \ln(3) - \ln(2))

This simplifies to:

(2\ln(2) - \ln(2) + \ln(3))

Combining the logarithms, we get the final result:

(\ln(2^2) + \ln(3) = \ln(4) + \ln(3) = \ln(12))