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Integration of Rational Functions: The Method of Partial Fractions solved questions
Integration of Rational Functions: The Method of Partial Fractions solved questions
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CHAPTER 30Integration 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
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