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

Feb 1, 2016

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

#### Explanation:

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

The method of partial fractions involves splitting the denominator into its factors. Lets ignore the integral for now. We are looking for two values, $A$ and $B$ such that;

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

Now we need to combine $A$ and $B$ over a common denominator.

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

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

We can cancel the denominator on both sides, leaving;

$A \left(x + 1\right) + B \left(x - 2\right) = 3$

Now we can solve for $A$ and $B$ by selectively choosing our values of $x$ so that the terms in parenthesis are $0$. Let $x = - 1$.

$A {\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(- 1 + 1\right)}}}}^{0} + B \left(- 1 - 2\right) = 3$

$- 3 B = 3$

$B = - 1$

Let $x = 2$.

$A \left(2 + 1\right) + B {\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(2 - 2\right)}}}}^{0} = 3$

$3 A = 3$

$A = 1$

Now that we have our values for $A$ and $B$ we can split apart our integral by plugging them in.

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

$= \int \frac{1}{x - 2} \mathrm{dx} - \int \frac{1}{x + 1} \mathrm{dx}$

Now we have two integrals that we can solve using substitution. Let ${u}_{1} = x - 2$ for the first integral and ${u}_{2} = x + 1$ for the second.

$\int \frac{1}{u} _ 1 {\mathrm{du}}_{1} - \int \frac{1}{u} _ 2 {\mathrm{du}}_{2}$

$\ln \left({u}_{1}\right) - \ln \left({u}_{2}\right) + C$

$\ln \left(\frac{{u}_{1}}{{u}_{2}}\right) + C$

Plug in the values for ${u}_{1}$ and ${u}_{2}$.

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