Algebra and Partial Fractions - Solutions Math 125
Integration of rational functions is mostly a matter of algebraic manipulation. In this worksheet we
shall work through some examples of the necessary techniques.
1a Consider the rational function f(x)=2x3−4x2−5x+3
x2−2x−3. Use long division to get a quotient
and a remainder, then write f(x) =quotient + (remainder/divisor).
f(x)=2x+x+3
x2−2x−3
1b Now consider the expression x+3
x2−2x−3. Factor the denominator into two linear terms.
x+3
(x+1)(x−3)
1c We wish to write x+3
(x−3)(x+1) as a sum A
x−3+B
x+1. Let’s find Aand B. Set the two
expressions equal and clear denominators (that is, multiply through by (x−3)(x+ 1) and cancel
(x−3)’s and (x+ 1)’s as much as possible). Plug in x= 3 and solve for A. Use the same idea to
find B. Check your work by adding the two fractions together.
x+3=A(x+1)+B(x−3)
If x=3,weget6=4Aso A=3
2.
If x=−1,weget2=−4Bso B=−1
2.
Check that 3/2
x−3+−1/2
x+1 =x+3
(x−3)(x+1).
1d Now use the results of Problems 1a, 1b, and 1c to compute !2x3−4x2−5x+3
x2−2x−3dx.
Putting it all together, f(x)=2x+3/2
x−3−1/2
x+1 so
!f(x)dx =!2x+3/2
x−3−1/2
x+1dx =x2+3
2ln |x−3|−1
2ln |x+1|+C.
1e Some of the terms in the answer to Problem 1d involve logarithms. Combine those terms into
a single term of the form ln(some function of x).
!f(x)dx =x2+ln"
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