2.5: Partial Fractions
( \newcommand{\kernel}{\mathrm{null}\,}\)
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
(x−r)n
or
x2+Ax+B.
Consider the rational function
P(x)=3x+2x2−1=3x+2(x−1)(x+1).
We want to write it in the form
3x+2(x−1)(x+1)=Ax−1+Bx+1.
To do this we need to solve for A and B. Multiplying by the common denominator
(x−1)(x+1)
we have
3x+2=A(x+1)+B(x−1).
Now let x=1
5=2A+0
A=52.
Now let x=−1
−1=−2B
B=12.
Hence we can write
3x+2x2−1=52x−1+12x+1.
This is called the partial fraction decomposition of P(x).
Find the Partial Fraction Decomposition of
P(x)=3x2+4x+7x3−2x2+x=3x2+4x+7x(x−1)2.
We write
3x2+4x+7x(x−1)2=Ax−1+B(x−1)2+Cx.
Multiplying by the common denominator, we have
A(x)(x−1)+Bx+C(x−1)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(x−1)2=−4x−1+14(x−1)2+7x.
Evaluate
∫x−2x(x2+1)dx.
Solution
We write
x2−2x(x2+1)=Ax+Bx+Cx2+1.
Multiplying by the common denominator, we have
A(x2+1)+(Bx+C)x=x2−2.
Let x=0
A=−2.
Hence
(Bx+C)x=x2−2+2x2+2=3x2.
So that
Bx2+Cx=3x2.
We see that B=3 and C=0
Hence
∫x2−2x(x2+1)dx=∫(−2x+3xx2+1)dx=−2∫1xdx+∫3xx2+1dxu=x2+1,du=2xdx=−2ln|x|+32∫duudx=−2ln|x|+32ln(x2+1)+C.
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(30,000−P).
This is a separable differential equation. Separating gives
dPP(30,000−P)=kdt.
Now integrate both sides. The right hand side is
kt+C.
To integrate the right hand side, use partial fractions:
1P(30,000−P)=AP+B30,000−P.
Multiplying by the common denominator:
1=A(30,000−P)+BP
P=0: 1=30,000A or A=130,000
P=30,000: 1=30,000B or B=130,000.
Now integrate to get
130,000lnP−130,000ln(30,000−P)=kt+C
or
lnP30,000−P=at+b.
When t=0, P=15,000, so
b=0.
When t=20, P=20,000, so
20a=ln2.
or
a=ln220
hence
lnP30,000−P=tln220.
Finally
ln5=tln220
or
t=20ln5ln2=46.4.
The population will reach 2500 in the year 2026.
Find:
- ∫x−2x2+3x−4dx
- ∫x(x+1)(x2+4)dx
- ∫2x+3x2(x−3)dx
- ∫6x+13x2+x−2dx
Contributors and Attributions
- Larry Green (Lake Tahoe Community College)
Integrated by Justin Marshall.