# How do you integrate int (x-1)/(3x^2-14x+15) using partial fractions?

Dec 4, 2016

The answer is =-1/6ln(∣3x-5∣)+1/2ln(∣x-3∣)+C

#### Explanation:

Let's factorise the denominator

$3 {x}^{2} - 14 x + 15 = \left(3 x - 5\right) \left(x - 3\right)$

So, the decomposition into partial fractions are

$\frac{x - 1}{3 {x}^{2} - 14 x + 15} = \frac{x - 1}{\left(3 x - 5\right) \left(x - 3\right)} = \frac{A}{3 x - 5} + \frac{B}{x - 3}$

$= \frac{A \left(x - 3\right) + B \left(3 x - 5\right)}{\left(3 x - 5\right) \left(x - 3\right)}$

So, $\left(x - 1\right) = A \left(x - 3\right) + B \left(3 x - 5\right)$

Let $x = 3$, $\implies$, $2 = 4 B$, $\implies$, $B = \frac{1}{2}$

Let $x = \frac{5}{3}$, $\implies$, $\frac{2}{3} = - 4 \frac{A}{3}$, $\implies$,$A = - \frac{1}{2}$

so,

$\frac{x - 1}{3 {x}^{2} - 14 x + 15} = \frac{- \frac{1}{2}}{3 x - 5} + \frac{\frac{1}{2}}{x - 3}$

$\int \frac{\left(x - 1\right) \mathrm{dx}}{3 {x}^{2} - 14 x + 15} = \int \frac{\left(- \frac{1}{2}\right) \mathrm{dx}}{3 x - 5} + \int \frac{\left(\frac{1}{2}\right) \mathrm{dx}}{x - 3}$

=-1/2*ln(∣3x-5∣)/3+1/2*ln(∣x-3∣)+C

=-1/6*ln(∣3x-5∣)+1/2*ln(∣x-3∣)+C