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

Dec 13, 2016

${\int}_{3}^{4} \frac{{x}^{2} - 3 x + 6}{x \left(x - 2\right) \left(x - 1\right)} \mathrm{dx} = 12 \ln 2 - 7 \ln 3$

#### Explanation:

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

$\setminus \setminus \setminus \setminus \setminus \frac{{x}^{2} - 3 x + 6}{x \left(x - 2\right) \left(x - 1\right)} \equiv \frac{A}{x} + \frac{B}{x - 2} + \frac{C}{x - 1}$
$\therefore \frac{{x}^{2} - 3 x + 6}{x \left(x - 2\right) \left(x - 1\right)} = \frac{A \left(x - 2\right) \left(x - 1\right) + B x \left(x - 1\right) + C x \left(x - 2\right)}{x \left(x - 2\right) \left(x - 1\right)}$

And so:

$\therefore {x}^{2} - 3 x + 6 = A \left(x - 2\right) \left(x - 1\right) + B x \left(x - 1\right) + C x \left(x - 2\right)$

We can easily find the constants $A , B , C$ by setting $x$ to the specific values that make the denominator 0 or by comparing coefficients. (With practice this can be done instantly using the "Cover Up" method).

Put $x = 0 \implies 6 = A \left(- 2\right) \left(- 1\right) \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \implies A = 3$
Put $x = 2 \implies 4 - 6 + 6 = B \left(2\right) \left(1\right) \setminus \setminus \setminus \setminus \setminus \implies B = 2$
Put $x = 1 \implies 1 - 3 + 6 = C \left(1\right) \left(- 1\right) \implies \setminus C = - 4$

So the integral can be written as:
${\int}_{3}^{4} \frac{{x}^{2} - 3 x + 6}{x \left(x - 2\right) \left(x - 1\right)} \mathrm{dx} = {\int}_{3}^{4} \frac{3}{x} + \frac{2}{x - 2} - \frac{4}{x - 1} \mathrm{dx}$

Which we can now integrate as we know that

$\frac{d}{\mathrm{dx}} \ln \left(a x + b\right) = \frac{a}{a x + b}$

Hence,

${\int}_{3}^{4} \frac{{x}^{2} - 3 x + 6}{x \left(x - 2\right) \left(x - 1\right)} \mathrm{dx} = {\left[3 \ln x + 2 \ln | x - 2 | - 4 \ln | x - 1 |\right]}_{3}^{4}$
${\int}_{3}^{4} \frac{{x}^{2} - 3 x + 6}{x \left(x - 2\right) \left(x - 1\right)} \mathrm{dx} = \left(3 \ln 4 + 2 \ln 2 - 4 \ln 3\right) - \left(3 \ln 3 + 2 \ln 1 - 4 \ln 2\right)$
${\int}_{3}^{4} \frac{{x}^{2} - 3 x + 6}{x \left(x - 2\right) \left(x - 1\right)} \mathrm{dx} = 3 \ln 4 + 2 \ln 2 - 4 \ln 3 - 3 \ln 3 - 0 + 4 \ln 2$
${\int}_{3}^{4} \frac{{x}^{2} - 3 x + 6}{x \left(x - 2\right) \left(x - 1\right)} \mathrm{dx} = 3 \ln 4 + 6 \ln 2 - 7 \ln 3$
${\int}_{3}^{4} \frac{{x}^{2} - 3 x + 6}{x \left(x - 2\right) \left(x - 1\right)} \mathrm{dx} = \left(2 \cdot 3\right) \ln 2 + 6 \ln 2 - 7 \ln 3$
${\int}_{3}^{4} \frac{{x}^{2} - 3 x + 6}{x \left(x - 2\right) \left(x - 1\right)} \mathrm{dx} = 12 \ln 2 - 7 \ln 3$
${\int}_{3}^{4} \frac{{x}^{2} - 3 x + 6}{x \left(x - 2\right) \left(x - 1\right)} \mathrm{dx} \approx 0.627480146042576$