Chapter 48: Partial Fractions

Partial Fractions (8.5)(\oint 8.5)

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

Example - 11x2=1(1x)(1+x)=A1x+B1+x\frac{1}{1-x^{2}}=\frac{1}{(1-x)(1+x)}=\frac{A}{1-x}+\frac{B}{1+x}

Solve for A and B

A1x+B1+x=1(1x)(1+x)A(1+x)+B(1x)=1\frac{A}{1-x}+\frac{B}{1+x}=\frac{1}{(1-x)(1+x)}\Rightarrow A(1+x)+B(1-x)=1

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


We obtain A=B=±A=B=\pm 

Thus, 11x2=121x+121+x\frac{1}{1-x^{2}}=\frac{\frac{1}{2}}{1-x}+\frac{\frac{1}{2}}{1+x}

Example - 11x2dx=12ln(x1)+12lnx+1+C\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

1(x2)(x+3)=Ax2+Bx+3\frac{1}{(x-2)(x+3)}=\frac{A}{x-2}+\frac{B}{x+3}

A(x+3)+B(x2)=1\Rightarrow A(x+3)+B(x-2)=1

(A+B)x+(3A2B)=0x+1\Rightarrow (A+B)x +(3A-2B)=0x+1

{3A2B=1A+B=0\Rightarrow \left\{\begin{matrix} 3A-2B=1\\A+B = 0 \end{matrix}\right. A=15A=\frac{1}{5} B=15B=-\frac{1}{5}

Using this decomposition we get

1x2+x6dx=15x215x+3dx\int \frac{1}{x^{2}+x-6}dx=\int \frac{\frac{1}{5}}{x-2}-\frac{\frac{1}{5}}{x+3}dx

=15lnx215lnx+3+C=\frac{1}{5}ln|x-2|-\frac{1}{5}ln|x+3|+C


Example - Express

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

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

6x+7=A(x+2)+BA=66x+7=A(x+2)+B \Rightarrow A=6 and B=5B=-5

Thus, 6x+7(x+2)2=6x+25(x+2)2\frac{6x+7}{(x+2)^{2}}=\frac{6}{x+2}-\frac{5}{(x+2)^{2}}


Discuss - Express

5x+4(x1)3=Ax1+B(x1)2+C(x1)3\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(x1)2+B(x1)+CA=05x+4 = A(x-1)^{2} +B(x-1)+C \Rightarrow A=0 B=5B=5 C=1C=1

Thus, 5x+4(x1)3=5(x1)2+1(x1)3\frac{5x+4}{(x-1)^{3}}=\frac{5}{(x-1)^{2}}+\frac{1}{(x-1)^{3}}


Example - Express

x2+4x+1(x1)(x+1)(x+3)=Ax1+Bx+1+Cx+3\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

x2+4x+1=A(x=1)(x+3)+B(x1)(x+3)+C(x1)(x+1)x^{2}+4x+1 = A (x=1)(x+3)+B(x-1)(x+3)+C(x-1)(x+1)

=A(x2+4x+3)+B(x2+2x3)+C(x2+0x1)=A(x^{2}+4x+3)+B(x^{2}+2x-3)+C(x^{2}+0x-1)

=(A+B+C)x2+(4A+2B)x+(3A+3BC)=(A+B+C)x^{2}+(4A+2B)x+(3A+3B-C)

{A+B+C=1(1)4A+2B=4(2)3A3BC=1(3)\left\{\begin{matrix} A+B+C = 1 & (1)\\ 4A +2B =4 &(2) \\3A-3B-C=1 &(3) \end{matrix}\right.


We obtain A=34A=\frac{3}{4} B=12B=\frac{1}{2} C=14C=-\frac{1}{4}

Method - If we have a repeated linear factor

ax+b(xc)n=A1x3+A2(xc)2+...+An(xc)n\frac{ax+b}{(x-c)^{n}}=\frac{A_{1}}{x-3}+\frac{A_{2}}{(x-c)^{2}}+... +\frac{An}{(x-c)^{n}}


Example - Express

x+4x(x2+1)=Ax+Bx2+1+Cx\frac{x+4}{x(x^{2}+1)}=\frac{Ax+B}{x^{2}+1}+\frac{C}{x}

Clear fractions

x+4=(Ax+B)x+C(x2+1)x+4 = (Ax+B)x+C(x^{2}+1)

=(A+C)x2+Bx+C=(A+C)x^{2}+Bx+C

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


We obtain x+4x(x2+1)=4x+1x2+1+4x\frac{x+4}{x(x^{2}+1)}=\frac{-4x+1}{x^{2}+1}+\frac{4}{x}


Example - Express

8x2+8x+2(4x2+1)2=Ax+B4x2+1+Cx+D(4x2+1)2\frac{8x^{2}+8x+2}{(4x^{2}+1)^{2}}=\frac{Ax+B}{4x^{2}+1}+\frac{Cx+D}{(4x^{2}+1)^{2}}

Clear fractions

8x2+8x+2=(Ax+B)(4x2+1)+(Cx+D)8x^{2}+8x +2 =(Ax +B)(4x^{2}+1)+(Cx+D)

=4Ax3+Ax+4Bx2+B+(x+D)=4Ax^{3}+Ax+4Bx^{2}+B+(x+D)

=4Ax3+4Bx2+(A+C)x+D=4Ax^{3}+4Bx^{2}+(A+C)x+D

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


Note Created by
Is this note helpful?
Give kudos to your peers!
00
Wanna make this note your own?
Fork this Note
17 Views