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2.5: Partial Fractions

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The Fundamental Theorem of Algebra

The fundamental theorem of algebra states that if P(x) is a polynomial of degree n then P(x) can be factored into linear factors over the complex numbers. Furthermore, P(x) can be factored over the real numbers as a product of linear and quadratic terms and any rational function can be split up as a sum of rational function with denominators of the form

(xr)n

or

x2+Ax+B.

Example 2.5.1: Partial Fractions

Consider the rational function

P(x)=3x+2x21=3x+2(x1)(x+1).

We want to write it in the form

3x+2(x1)(x+1)=Ax1+Bx+1.

To do this we need to solve for A and B. Multiplying by the common denominator

(x1)(x+1)

we have

3x+2=A(x+1)+B(x1).

Now let x=1

5=2A+0

A=52.

Now let x=1

1=2B

B=12.

Hence we can write

3x+2x21=52x1+12x+1.

This is called the partial fraction decomposition of P(x).

Example 2.5.2

Find the Partial Fraction Decomposition of

P(x)=3x2+4x+7x32x2+x=3x2+4x+7x(x1)2.

We write

3x2+4x+7x(x1)2=Ax1+B(x1)2+Cx.

Multiplying by the common denominator, we have

A(x)(x1)+Bx+C(x1)2=3x2+4x+7.

Let x=0:

C=7.

Let x=1:

We have

B=14.

Now look at the highest degree coefficient:

Ax2+Cx2=3x2.

Dividing by x2 and substituting C=7:

A+7=3, A=4.

We conclude that

3x2+4x+7x(x1)2=4x1+14(x1)2+7x.

Example 2.5.3

Evaluate

x2x(x2+1)dx.

Solution

We write

x22x(x2+1)=Ax+Bx+Cx2+1.

Multiplying by the common denominator, we have

A(x2+1)+(Bx+C)x=x22.

Let x=0

A=2.

Hence

(Bx+C)x=x22+2x2+2=3x2.

So that

Bx2+Cx=3x2.

We see that B=3 and C=0

Hence

x22x(x2+1)dx=(2x+3xx2+1)dx=21xdx+3xx2+1dxu=x2+1,du=2xdx=2ln|x|+32duudx=2ln|x|+32ln(x2+1)+C.

Example 2.5.4

It has been said that the total population of South Lake Tahoe should never exceed 30,000 people. If in 1980 the population was 15,000 and in 2000 it was 20,000, when will the population reach 25,000?

Solution

We make the assumption that the rate of increase of the population is proportional to the product of the current population and 30,000 minus the current population. That is:

dPdt=kP(30,000P).

This is a separable differential equation. Separating gives

dPP(30,000P)=kdt.

Now integrate both sides. The right hand side is

kt+C.

To integrate the right hand side, use partial fractions:

1P(30,000P)=AP+B30,000P.

Multiplying by the common denominator:

1=A(30,000P)+BP

P=0: 1=30,000A or A=130,000

P=30,000: 1=30,000B or B=130,000.

Now integrate to get

130,000lnP130,000ln(30,000P)=kt+C

or

lnP30,000P=at+b.

When t=0, P=15,000, so

b=0.

When t=20, P=20,000, so

20a=ln2.

or

a=ln220

hence

lnP30,000P=tln220.

Finally

ln5=tln220

or

t=20ln5ln2=46.4.

The population will reach 2500 in the year 2026.

Exercises 2.5.1

Find:

  1. x2x2+3x4dx
  2. x(x+1)(x2+4)dx
  3. 2x+3x2(x3)dx
  4. 6x+13x2+x2dx

Contributors and Attributions


This page titled 2.5: Partial Fractions is shared under a not declared license and was authored, remixed, and/or curated by Larry Green.

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