# How do you find int (x - 5x^2) / ((x+2)(x-1)(x-3)) dx using partial fractions?

Jan 29, 2017

The answer is $= - \frac{22}{15} \ln \left(| x + 2 |\right) + \frac{2}{3} \ln \left(| x - 1 |\right) - \frac{21}{5} \ln \left(| x - 3 |\right) + C$

#### Explanation:

Let's work out the decomposition into partial fractions

$\frac{x - 5 {x}^{2}}{\left(x + 2\right) \left(x - 1\right) \left(x - 3\right)}$

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

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

The denominators are the same, we can compare the numerators

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

Let $x = - 2$, $\implies$, $- 22 = 15 A$, $\implies$, $A = - \frac{22}{15}$

Let $x = 1$, $\implies$, $- 4 = - 6 B$, $\implies$, $B = \frac{2}{3}$

Let $x = 3$, $\implies$, $- 42 = 10 C$, $\implies$, $C = - \frac{21}{5}$

Therefore,

$\frac{x - 5 {x}^{2}}{\left(x + 2\right) \left(x - 1\right) \left(x - 3\right)} = \frac{- \frac{22}{15}}{x + 2} + \frac{\frac{2}{3}}{x - 1} + \frac{- \frac{21}{5}}{x - 3}$

So,

$\int \frac{\left(x - 5 {x}^{2}\right) \mathrm{dx}}{\left(x + 2\right) \left(x - 1\right) \left(x - 3\right)} = - \frac{22}{15} \int \frac{\mathrm{dx}}{x + 2} + \frac{2}{3} \int \frac{\mathrm{dx}}{x - 1} - \frac{21}{5} \int \frac{\mathrm{dx}}{x - 3}$

$= - \frac{22}{15} \ln \left(| x + 2 |\right) + \frac{2}{3} \ln \left(| x - 1 |\right) - \frac{21}{5} \ln \left(| x - 3 |\right) + C$