CHAPTER 30
Integration of Rational Functions:
The Method of Partial Fractions
In Problems 30.1-30.21, evaluate the indicated antiderivative.
30.1
Clear the denominators by multiplying both sides by (x -
3)(x + 3): l = AO + 3)+B(.r-3). Let x = 3. Then 1 = 6A, A = i . Let x = -3. Then 1 = -6B,
* = - J . So l/(* 2 -9)=J[l/(*-3)]-J[l/(^ + 3)]. Hence, = 4 l n | j c - 3 | -4In |x+3|+C=
iln|(JC-3)/(x + 3)| + C.
30.2
Then x = A(x + 3) + B(x + 2). Let j c = - 3 . Then - 3 = - B , B = 3.
Let A: = -2. Then -2 = y4. So Hence,
3 2
-2 In |JT + 2| + 3 In |jc + 3| + C = In |(x + 3) /(x + 2) | + C.
30.3
Since the degree of the numerator is at least as great as that of the denominator, carry out the long division,
obtaining Thus, But
Then jc + 1 = A(x -2) + B(;t + 2). Let jc = 2. Then 3 = 4B,
B=3/4. Let x = -2. Then -1 = -4A, A = \. Thus, and
In In ln|(jc + 2)(A:-2) 3 | + C.
In Hence, the complete answer
3
is In K* + 2)(x - 2) | -t- C.
30.4
Then 2^2 + 1 = A(A: -2)(x - 3) + B(^ - l)(x - 3) +
C(*-!)(*-2). Let ^ = 1. Then 3 = 2/1, X=§. Let x = 2. Then 9=-B, B = -9. Let
x = 3. Then 19 = 2C, C=f. Thus, Hence,
dr = § In |x - 1| - 9 In |^ - 2| + ¥ In |^ - 3| + C, = ^ l n + C,.
30.5
We must factor the denominator. Clearly x = 1 is a root. Dividing the denominator by x -1
we obtain x2-2x-3 = (x-3)(x+l). Hence, the denominator is (x — l)(x — 3)(jc + 1).
Then x2 - 4 = A(x - 3)(x + 1) +fi(;c- l)(x + 1) + C(x -
l)(x-3). Let JK = I. Then -3=-4A, A=\. Let jt = 3. Then 5 = 8B, B = | . Let x=-l.
245
, 246 CHAPTER 30
Then -3 = 8C, C=-|. Thus,
In In In In
30.6
So jc3 + 1 = A(x + 3)(x + 2)(x -l) + Bx(x + 2)(x -
1) + CA-(A- + 3)(x - 1) + Dx(x + 3)(x + 2). Let x = 0. Then l = -6A, A = -k. Let * = -3. Then
-26=-12B, S = f . Let x = -2. Then -7 = 6C, C = -1. Let x = l. Then 2 = 12D, D = £ .
Thus, and
In In In In In
30.7
x 4 - I3x2 + 36 = (x2 - 9)(x2 - 4) = (x - 3)(x + 3)(x + 2)(x - 2). Let
Then x = A(x + 3)(x + 2)(x - 2) + B(x - 3)0 + 2)(x - 2) + C(x - 3)(x +
3)(x -2) + D(x - 3)(x + 3)(x + 2). Let x = 3. Then 3 = 30.4. A=&. Let x = -3. Then -3 =
-305. B=-k. Let x = -2. Then -2 = 20C, C =.-•&. Let x = 2. Then 2=-20D, D = -^.
So and
In In in In In
30.8
Multiply both sides by x\x + l), obtaining x-5 = Ax(x + 1) + B(x +
2
1) + Cx . Let x = 0. Then -5 = B. Let * = -!. Then -6 = C. To find A, compare coefficients of
2
x on both sides of the equation: 0=A + C, A = -C = 6. Thus, and
In In In
30.9
Then 2x = A(x - 2)(x + 2) + B(x + 2)+ C(x - 2)2. Let
x = 2. Then 4 = 4fl, 5=1. Let x = -2. Then -4=16C. C = - i . To find A, equate coefficients of x2:
0 = A+ C, A = -C=\. Thus, and
In In In
30.10
The benefits of buying summaries with Stuvia:
Guaranteed quality through customer reviews
Stuvia customers have reviewed more than 700,000 summaries. This how you know that you are buying the best documents.
Quick and easy check-out
You can quickly pay through credit card for the summaries. There is no membership needed.
Focus on what matters
Your fellow students write the study notes themselves, which is why the documents are always reliable and up-to-date. This ensures you quickly get to the core!
Frequently asked questions
What do I get when I buy this document?
You get a PDF, available immediately after your purchase. The purchased document is accessible anytime, anywhere and indefinitely through your profile.
Satisfaction guarantee: how does it work?
Our satisfaction guarantee ensures that you always find a study document that suits you well. You fill out a form, and our customer service team takes care of the rest.
Who am I buying these notes from?
Stuvia is a marketplace, so you are not buying this document from us, but from seller jureloqoo. Stuvia facilitates payment to the seller.
Will I be stuck with a subscription?
No, you only buy these notes for £6.47. You're not tied to anything after your purchase.