# Chapter 48: Partial Fractions

### Partial Fractions ﻿$(\oint 8.5)$﻿

When integrating ratios of polynomials we need to use some special techniques. One maneuver is to break up a fraction.

Example - ﻿$\frac{1}{1-x^{2}}=\frac{1}{(1-x)(1+x)}=\frac{A}{1-x}+\frac{B}{1+x}$﻿

Solve for A and B

﻿$\frac{A}{1-x}+\frac{B}{1+x}=\frac{1}{(1-x)(1+x)}\Rightarrow A(1+x)+B(1-x)=1$﻿

﻿$(A+B)+(A-B)x=1+0x \Rightarrow \left\{\begin{matrix} A+B=1\\A-B=0 \end{matrix}\right.$﻿

We obtain ﻿$A=B=\pm$﻿

Thus, ﻿$\frac{1}{1-x^{2}}=\frac{\frac{1}{2}}{1-x}+\frac{\frac{1}{2}}{1+x}$﻿

Example - ﻿$\int \frac{1}{1-x^{2}}dx=\frac{1}{2}ln(x-1)+\frac{1}{2}ln|x+1|+C$﻿

Discuss - Find A and B so that

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

﻿$\Rightarrow A(x+3)+B(x-2)=1$﻿

﻿$\Rightarrow (A+B)x +(3A-2B)=0x+1$﻿

﻿$\Rightarrow \left\{\begin{matrix} 3A-2B=1\\A+B = 0 \end{matrix}\right.$﻿ ﻿$A=\frac{1}{5}$﻿ ﻿$B=-\frac{1}{5}$﻿

Using this decomposition we get

﻿$\int \frac{1}{x^{2}+x-6}dx=\int \frac{\frac{1}{5}}{x-2}-\frac{\frac{1}{5}}{x+3}dx$﻿

﻿$=\frac{1}{5}ln|x-2|-\frac{1}{5}ln|x+3|+C$﻿

Example - Express

﻿$\frac{6x+7}{(x+2)^{2}}=\frac{A}{x+2}+\frac{B}{(x+2)^{2}}$﻿

Clear fractions by multiplying ﻿$(x+2)^{2}$﻿

﻿$6x+7=A(x+2)+B \Rightarrow A=6$﻿ and ﻿$B=-5$﻿

Thus, ﻿$\frac{6x+7}{(x+2)^{2}}=\frac{6}{x+2}-\frac{5}{(x+2)^{2}}$﻿

Discuss - Express

﻿$\frac{5x+4}{(x-1)^{3}}=\frac{A}{x-1}+\frac{B}{(x-1)^{2}}+\frac{C}{(x-1)^{3}}$﻿

Clear Fractions

﻿$5x+4 = A(x-1)^{2} +B(x-1)+C \Rightarrow A=0$﻿ ﻿$B=5$﻿ ﻿$C=1$﻿

Thus, ﻿$\frac{5x+4}{(x-1)^{3}}=\frac{5}{(x-1)^{2}}+\frac{1}{(x-1)^{3}}$﻿

Example - Express

﻿$\frac{x^{2}+4x+1}{(x-1)(x+1)(x+3)}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{x+3}$﻿

Clear fractions

﻿$x^{2}+4x+1 = A (x=1)(x+3)+B(x-1)(x+3)+C(x-1)(x+1)$﻿

﻿$=A(x^{2}+4x+3)+B(x^{2}+2x-3)+C(x^{2}+0x-1)$﻿

﻿$=(A+B+C)x^{2}+(4A+2B)x+(3A+3B-C)$﻿

﻿$\left\{\begin{matrix} A+B+C = 1 & (1)\\ 4A +2B =4 &(2) \\3A-3B-C=1 &(3) \end{matrix}\right.$﻿

We obtain ﻿$A=\frac{3}{4}$﻿ ﻿$B=\frac{1}{2}$﻿ ﻿$C=-\frac{1}{4}$﻿

Method - If we have a repeated linear factor

﻿$\frac{ax+b}{(x-c)^{n}}=\frac{A_{1}}{x-3}+\frac{A_{2}}{(x-c)^{2}}+... +\frac{An}{(x-c)^{n}}$﻿

Example - Express

﻿$\frac{x+4}{x(x^{2}+1)}=\frac{Ax+B}{x^{2}+1}+\frac{C}{x}$﻿

Clear fractions

﻿$x+4 = (Ax+B)x+C(x^{2}+1)$﻿

﻿$=(A+C)x^{2}+Bx+C$﻿

﻿$\left\{\begin{matrix} A+C & =0 \\B &=1 \\ C & =4 \end{matrix}\right.$﻿ Thus, ﻿$A=-4$﻿ ﻿$B=1$﻿ ﻿$C=4$﻿

We obtain ﻿$\frac{x+4}{x(x^{2}+1)}=\frac{-4x+1}{x^{2}+1}+\frac{4}{x}$﻿

Example - Express

﻿$\frac{8x^{2}+8x+2}{(4x^{2}+1)^{2}}=\frac{Ax+B}{4x^{2}+1}+\frac{Cx+D}{(4x^{2}+1)^{2}}$﻿

Clear fractions

﻿$8x^{2}+8x +2 =(Ax +B)(4x^{2}+1)+(Cx+D)$﻿

﻿$=4Ax^{3}+Ax+4Bx^{2}+B+(x+D)$﻿

﻿$=4Ax^{3}+4Bx^{2}+(A+C)x+D$﻿

Thus, ﻿$A=0,$﻿ ﻿$B=4,$﻿ ﻿$C=8,$﻿ ﻿$D=2$﻿

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