# Use partial fraction to find the power series of the function \frac {3}{(x+2)(x-1)}

## Question:

Use partial fraction to find the power series of the function {eq}\frac {3}{(x+2)(x-1)} {/eq}

## Simple fractions:

In the decomposition into simple fractions of a rational function, four cases are distinguished:

1. Simple real roots (different linear factors).

2. Multiple real roots. (Repeated linear factors).

3. Simple complex roots.

4. Multiple complex roots.

Part 1.

Resolve into partial fractions.

{eq}\displaystyle \frac {3}{(x+2)(x-1)}= \frac {A}{x-1} +\frac {B}{x+2}\\ \displaystyle \frac {3}{(x+2)(x-1)}= \frac {A(x+2)+B(x-1)}{(x-1)(x+2)}\\ \displaystyle \frac {3}{(x+2)(x-1)}= \frac {Ax+2A+Bx-B}{(x-1)(x+2)}\\ \displaystyle \frac {3}{(x+2)(x-1)}= \frac {x(A+B)+2A-B}{(x-1)(x+2)}\\ \displaystyle A+B=0 \\ \displaystyle 2A-B=3\\ \displaystyle A=1 \\ \displaystyle B=-1 \\ {/eq}

The fraction {eq}\frac {3}{(x+2)(x-1)} {/eq} into partial fractions are:

{eq}\displaystyle \Longrightarrow \boxed{ \frac {1}{x-1} -\frac {1}{x+2}}\\ {/eq}

Pat 2.

Calculate the integral

{eq}\displaystyle I = \int \frac {3}{(x+2)(x-1)} \, dx \\ \displaystyle I = \int \frac {1}{x-1} -\frac {1}{x+2} \, dx\\ \displaystyle I = \ln |x-1|- \ln |x+2| +C\\ {/eq}

The result of the integral applying partial fraction method is:

{eq}\displaystyle \Longrightarrow \boxed{ I = \ln |x-1|-\ln |x+2| +C}\\ {/eq} 