AP Calculus BC Practice Quiz: Integrating Using Linear Partial Fractions (BC ONLY)

Correct Answer: A

To find the integral of 1/((x+1)(x+2)), we first perform a partial fraction decomposition. We set up the equation: 1/((x+1)(x+2)) = A/(x+1) + B/(x+2). Multiplying both sides by (x+1)(x+2) gives 1 = A(x+2) + B(x+1). To solve for A, let x = -1, which gives 1 = A(-1+2), so A = 1. To solve for B, let x = -2, which gives 1 = B(-2+1), so B = -1. The integral becomes ∫(1/(x+1) - 1/(x+2)) dx. Integrating term by term gives ln|x+1| - ln|x+2| + C.

Correct Answer: B

First, factor the denominator: x² + x - 2 = (x+2)(x-1). Then, set up the partial fraction decomposition: (x+4)/((x+2)(x-1)) = A/(x+2) + B/(x-1). This leads to the equation x+4 = A(x-1) + B(x+2). If we let x = 1, we get 1+4 = B(1+2), so 5 = 3B, and B = 5/3. If we let x = -2, we get -2+4 = A(-2-1), so 2 = -3A, and A = -2/3. The integral is ∫ ((-2/3)/(x+2) + (5/3)/(x-1)) dx. Integrating gives -(2/3)ln|x+2| + (5/3)ln|x-1| + C.

Correct Answer: D

This question tests the setup of a partial fraction decomposition. We start with the equation (4x + 7) / ((x-2)(x+3)) = A/(x-2) + B/(x+3). Multiply by the common denominator to get 4x + 7 = A(x+3) + B(x-2). To solve for A, we can choose x=2 (the root of x-2) which gives 4(2) + 7 = A(2+3) + B(0), so 15 = 5A, which means A = 3. To solve for B, we can choose x=-3 (the root of x+3) which gives 4(-3) + 7 = A(0) + B(-3-2), so -5 = -5B, which means B = 1.

Correct Answer: A

First, factor the denominator: x² - x - 6 = (x-3)(x+2). Decompose the fraction: (5x-5)/((x-3)(x+2)) = A/(x-3) + B/(x+2). This gives 5x-5 = A(x+2) + B(x-3). For x=3, 10=5A so A=2. For x=-2, -15=-5B so B=3. The integral is ∫(2/(x-3) + 3/(x+2)) dx. The antiderivative is 2ln|x-3| + 3ln|x+2|. Evaluating from 0 to 1: [2ln|1-3| + 3ln|1+2|] - [2ln|0-3| + 3ln|0+2|] = [2ln(2) + 3ln(3)] - [2ln(3) + 3ln(2)] = 2ln(2) + 3ln(3) - 2ln(3) - 3ln(2) = ln(3) - ln(2) = ln(3/2).

Correct Answer: B

First, factor the denominator: x³ - x = x(x²-1) = x(x-1)(x+1). The partial fraction decomposition is A/x + B/(x-1) + C/(x+1). This gives 6 = A(x-1)(x+1) + Bx(x+1) + Cx(x-1). For x=0, 6 = A(-1), so A=-6. For x=1, 6 = B(1)(2), so B=3. For x=-1, 6 = C(-1)(-2), so C=3. The integral is ∫(-6/x + 3/(x-1) + 3/(x+1)) dx. This integrates to -6ln|x| + 3ln|x-1| + 3ln|x+1| + C. Using logarithm properties, 3ln|x-1| + 3ln|x+1| = 3(ln|x-1| + ln|x+1|) = 3ln|(x-1)(x+1)| = 3ln|x²-1|. So the final answer is -6ln|x| + 3ln|x²-1| + C.

Correct Answer: B

Factor the denominator: x² - x - 2 = (x-2)(x+1). Set up the partial fraction decomposition: 1/((x-2)(x+1)) = A/(x-2) + B/(x+1). This gives 1 = A(x+1) + B(x-2). For x=2, 1 = 3A, so A=1/3. For x=-1, 1 = -3B, so B=-1/3. The integral is ∫(1/3)/(x-2) - (1/3)/(x+1) dx. The antiderivative is (1/3)[ln|x-2| - ln|x+1|]. Evaluating from 3 to 4: (1/3)[(ln|4-2| - ln|4+1|) - (ln|3-2| - ln|3+1|)] = (1/3)[(ln(2) - ln(5)) - (ln(1) - ln(4))] = (1/3)[ln(2) - ln(5) + ln(4)] = (1/3)[ln(8) - ln(5)] = (1/3)ln(8/5).

Correct Answer: A

Factor the denominator: x² - 9 = (x-3)(x+3). The partial fraction decomposition is 4x/((x-3)(x+3)) = A/(x-3) + B/(x+3). This gives 4x = A(x+3) + B(x-3). For x=3, 12 = 6A, so A=2. For x=-3, -12 = -6B, so B=2. The integral is ∫(2/(x-3) + 2/(x+3)) dx. The antiderivative is 2ln|x-3| + 2ln|x+3|. This can be simplified to 2ln|(x-3)(x+3)| = 2ln|x²-9|. Now evaluate from 1 to 2: [2ln|x²-9|] from 1 to 2 = 2ln|2²-9| - 2ln|1²-9| = 2ln|-5| - 2ln|-8| = 2ln(5) - 2ln(8) = 2(ln(5) - ln(8)) = 2ln(5/8). The integral does not diverge as the function is continuous on the interval [1, 2].