# How do you integrate (2x-3)/(x^2-5x+6) using partial fractions?

Dec 3, 2016

#### Explanation:

Do partial fraction decomposition:

$\frac{2 x - 3}{\left(x - 2\right) \left(x - 3\right)} = \frac{A}{x - 2} + \frac{B}{x - 3}$

Multiply both sides by $\left(\left(x - 2\right) \left(x - 3\right)\right)$

$\left(2 x - 3\right) = A \left(x - 3\right) + B \left(x - 2\right)$

Solve for A by letting $x = 2$:

$\left(2 \left(2\right) - 3\right) = A \left(2 - 3\right)$

$- 1 = A$

Solve for B by letting $x = 3$:

$\left(2 \left(3\right) - 3\right) = B \left(3 - 2\right)$

3 = B

Check:

$- \frac{1}{x - 2} + \frac{3}{x - 3} =$

$- \frac{1}{x - 2} \frac{x - 3}{x - 3} + \frac{3}{x - 3} \frac{x - 2}{x - 2} =$

$\frac{- x + 3}{\left(x - 2\right) \left(x - 3\right)} + \frac{3 x - 6}{\left(x - 3\right) \left(x - 2\right)} =$

$\frac{2 x - 3}{\left(x - 3\right) \left(x - 2\right)}$

This checks.

$\int \frac{2 x - 3}{{x}^{2} - 5 x + 6} \mathrm{dx} = 3 \int \frac{1}{x - 3} \mathrm{dx} - \int \frac{1}{x - 2} \mathrm{dx}$

$\int \frac{2 x - 3}{{x}^{2} - 5 x + 6} \mathrm{dx} = 3 \ln | x - 3 | - \ln | x - 2 | + C$