# How do you integrate int 1/ [ (x-1)(x+2)(x-3)] dx using partial fractions?

Apr 10, 2017

$= - \frac{1}{6} \ln | x - 1 | + \frac{1}{15} \ln | x + 2 | + \frac{1}{10} \ln | x - 3 | + C$

#### Explanation:

Using partial fraction decomposition , break up the fraction into three different fractions added together:

$\int \frac{1}{\left(x - 1\right) \left(x + 2\right) \left(x - 3\right)} \mathrm{dx}$

$= \int \left(\frac{\textcolor{b l u e}{A}}{x - 1} + \frac{\textcolor{red}{B}}{x + 2} + \frac{\textcolor{g r e e n}{C}}{x - 3}\right) \mathrm{dx}$

$- - -$
In order to find the values of $\textcolor{b l u e}{A} , \textcolor{red}{B} , \mathmr{and} \textcolor{g r e e n}{C}$, use common denominators, and set it equal to the original fraction.

$\textcolor{b l u e}{A} \left(x + 2\right) \left(x - 3\right) + \textcolor{red}{B} \left(x - 1\right) \left(x - 3\right) + \textcolor{g r e e n}{C} \left(x - 1\right) \left(x + 2\right) = 1$

Plug in any value for $x$. The easiest numbers to plug in are those which make one of the factors zero, because many terms cancel:

$\text{If x=1: } \textcolor{b l u e}{A} \left(1 + 2\right) \left(1 - 3\right) = 1$
$\textcolor{b l u e}{A} = - \frac{1}{6}$

$\text{If x=-2: } \textcolor{red}{B} \left(- 2 - 1\right) \left(- 2 - 3\right) = 1$
$\textcolor{red}{B} = \frac{1}{15}$

$\text{If x=3: } \textcolor{g r e e n}{C} \left(3 - 1\right) \left(3 + 2\right) = 1$
$\textcolor{g r e e n}{C} = \frac{1}{10}$
$- - -$
The integral after plugging in values of $\textcolor{b l u e}{A} , \textcolor{red}{B} , \mathmr{and} \textcolor{g r e e n}{C}$ becomes:
$= \int \left(\frac{\textcolor{b l u e}{- \frac{1}{6}}}{x - 1} + \frac{\textcolor{red}{\frac{1}{15}}}{x + 2} + \frac{\textcolor{g r e e n}{\frac{1}{10}}}{x - 3}\right) \mathrm{dx}$

$= - \frac{1}{6} \ln | x - 1 | + \frac{1}{15} \ln | x + 2 | + \frac{1}{10} \ln | x - 3 | + C$