help The integral is found by first writing the partial fraction decomposition o

Question

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The integral is found by first writing the partial fraction decomposition of the integrand: 1 / x2 + 3x - 4 = 1 / (x + 4) (x - 1) = A / x + 4 + B / x - 1 and then computing the unknown constants A and B. What is the sum A + B of these constants? The integral is found by first writing the partial fraction decomposition of the integrand: x2 / (x - 3) (x + 2 )2T = A / x - 3 +B / x + 2 + C / (x + 2)2 The coefficient B turns out to be a fraction, B = p/q. What is the sum p + q of the numerator and denominator of this fraction?

Explanation / Answer

1. By comparison:

1 = A(x-1) + B(x+4)

So, 1 = x(A+B) + 4B - A

So, A+B = 0 and 4B - A = 1

Hence sum A+B is 0

2. by comparison:

A(x+2)^2 + B(x+2)(x-3) + C(x-3) = x^2

So,

A+B+C = 1

4A- B+C = 0

4A - 6B - 3C = 0

So, 5B+4C = 0

5B+3C = 4

So, C = -4

B = -16/5

So, p + q = -16 + 5 = -11

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