AP Calculus BC Quiz: Using Linear Partial Fractions

What this quiz covers

This quiz focuses on Using Linear Partial Fractions, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus BC.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $\frac{4}{3}\ln|x+2|+\frac{1}{3}\ln|x+5|+C$
2. $\frac{4}{3}\ln|x+2|-\frac{1}{3}\ln|x+5|+C$ (correct answer)
3. $-\frac{4}{3}\ln|x+2|+\frac{1}{3}\ln|x+5|+C$
4. $\frac{1}{3}\ln|x+2|-\frac{4}{3}\ln|x+5|+C$
5. $\frac{1}{3}\ln|x+2|+\frac{4}{3}\ln|x+5|+C$

Explanation: This problem involves using partial fraction decomposition for integration of a rational function with distinct linear factors. To decompose (x+6)/((x+2)(x+5)), write it as A/(x+2) + B/(x+5) and solve for A and B by multiplying through by the denominator, yielding A(x+5) + B(x+2) = x+6. Plugging in x=-2 gives A=4/3, and x=-5 gives B=-1/3. The integral then becomes (4/3) ∫ dx/(x+2) + (-1/3) ∫ dx/(x+5) = (4/3) ln|x+2| - (1/3) ln|x+5| + C. A tempting distractor like choice A ignores the negative sign from the decomposition, failing because it results in a mismatched numerator when verified. In general, for integrating rational functions with distinct linear denominators, decompose into partial fractions, solve the system by plugging in roots or equating coefficients, and integrate each term separately.

Question 2

1. $\frac{5}{2}\ln|t-2|+\frac{5}{2}\ln|t+2|+\frac{1}{4}\ln\left|\frac{t-2}{t+2}\right|+C$
2. $\frac{11}{4}\ln|t-2|+\frac{9}{4}\ln|t+2|+C$ (correct answer)
3. $\frac{9}{4}\ln|t-2|+\frac{11}{4}\ln|t+2|+C$
4. $\frac{5}{2}\ln|t^2-4|+\frac{1}{4}\ln|t^2-4|+C$
5. $\frac{5}{2}\ln|t-2|+\frac{1}{4}\ln|t+2|+C$

Explanation: The skill used here is partial fraction decomposition to integrate the rational function. Factor the denominator as t² - 4 = (t - 2)(t + 2), and express (5t + 1)/((t - 2)(t + 2)) as A/(t - 2) + B/(t + 2). Solving the equation 5t + 1 = A(t + 2) + B(t - 2) yields A = 11/4 and B = 9/4. Integrating term by term gives (11/4) ln|t - 2| + (9/4) ln|t + 2| + C. A tempting distractor like choice D combines the logs with incorrect coefficients, failing to account for the distinct constants from the decomposition. Always solve for the constants in partial fractions by substituting the roots or equating coefficients to ensure accurate integration of rational functions.

Question 3

1. $\frac{1}{2}\ln\left|\frac{x+1}{x+7}\right|+C$ (correct answer)
2. $\frac{1}{2}\ln\left|\frac{x+7}{x+1}\right|+C$
3. $\frac{1}{2}\ln|x+1|+\frac{1}{2}\ln|x+7|+C$
4. $\frac{3}{(x+1)(x+7)}+C$
5. $\ln|x+1| - \ln|x+7|+C$

Explanation: This problem requires the use of partial fraction decomposition to integrate a rational function. To evaluate ∫ 3 / ((x+1)(x+7)) dx, decompose the integrand as 3 / ((x+1)(x+7)) = A/(x+1) + B/(x+7). Solving for the coefficients gives A = 1/2 and B = -1/2. Integrating term by term yields (1/2) ln |x+1| - (1/2) ln |x+7| + C, which simplifies to (1/2) ln |(x+1)/(x+7)| + C. A tempting distractor is choice B, (1/2) ln |(x+7)/(x+1)| + C, but its derivative is -3 / ((x+1)(x+7)), the negative of the integrand. A transferable partial-fraction strategy is to multiply both sides by the common denominator and solve the resulting linear system for the coefficients.

Question 4

1. $2\ln|x|+3\ln|x+6|+C$ (correct answer)
2. $3\ln|x|+2\ln|x+6|+C$
3. $2\ln|x| - 3\ln|x+6|+C$
4. $3\ln|x+6| - 2\ln|x|+C$
5. $\frac{5x+12}{x(x+6)}+C$

Explanation: This problem requires the use of partial fraction decomposition to integrate a rational function. To evaluate ∫ (5x+12) / (x(x+6)) dx, decompose the integrand as (5x+12) / (x(x+6)) = A/x + B/(x+6). Solving for the coefficients gives A = 2 and B = 3. Integrating term by term yields 2 ln |x| + 3 ln |x+6| + C. A tempting distractor is choice C, 2 ln |x| - 3 ln |x+6| + C, but its derivative is (5x - 18) / (x(x+6)), not matching 5x+12. A transferable partial-fraction strategy is to multiply both sides by the common denominator and solve the resulting linear system for the coefficients.

Question 5

1. $\frac{1}{4}\ln|x-2|+\frac{1}{4}\ln|x+2|+C$
2. $\frac{1}{4}\ln\left|\frac{x-2}{x+2}\right|+C$ (correct answer)
3. $\frac{1}{2}\ln\left|\frac{x-2}{x+2}\right|+C$
4. $\frac{1}{4}\ln|x+2| - \frac{1}{4}\ln|x-2|+C$
5. $\frac{1}{(x-2)(x+2)}+C$

Explanation: This problem requires the use of partial fraction decomposition to integrate the rational function in an economics model. To decompose 1/((x-2)(x+2)), express it as A/(x-2) + B/(x+2), leading to 1 = A(x+2) + B(x-2). Solving gives A = 1/4 and B = -1/4 by substituting x=2 and x=-2. Integrating each term yields (1/4) ln|x-2| - (1/4) ln|x+2| + C, or equivalently (1/4) ln|(x-2)/(x+2)| + C. A tempting distractor is choice D, which separates with wrong signs, but this fails because it reverses the coefficient signs. In general, for partial fractions with distinct linear factors, solve the system for coefficients accurately and verify by recombining or differentiating the antiderivative.

Question 6

1. $\ln|x|+\ln|x+4|+C$
2. $\ln|x|+C$ (correct answer)
3. $\ln|x+4|+C$
4. $\ln\left|\frac{x}{x+4}\right|+C$
5. $\frac{x+4}{x(x+4)}+C$

Explanation: This problem requires the use of partial fraction decomposition to integrate the rational function in a fluid model. To decompose $\frac{x+4}{x(x+4)}$, express it as $\frac{A}{x} + \frac{B}{x+4}$, leading to $x+4 = A(x+4) + B x$. Solving gives $A = 1$ and $B = 0$ by substituting $x=0$ and $x=-4$, simplifying to $\frac{1}{x}$. Integrating yields $\ln|x| + C$. A tempting distractor is choice A, which includes an extra log, but this fails because the B coefficient is zero, so no second term. In general, for partial fractions with distinct linear factors, solve the system for coefficients accurately and verify by recombining or differentiating the antiderivative.

Question 7

1. $\frac{1}{2}\ln|x|+\frac{9}{2}\ln|x+2|+C$ (correct answer)
2. $\frac{1}{2}\ln|x| - \frac{9}{2}\ln|x+2|+C$
3. $\frac{9}{2}\ln|x|+\frac{1}{2}\ln|x+2|+C$
4. $\frac{5x+1}{x(x+2)}+C$
5. $\frac{9}{2}\ln|x+2| - \frac{1}{2}\ln|x|+C$

Explanation: This problem requires the use of partial fraction decomposition to integrate the rational function for probability density normalization. To decompose (5x+1)/(x(x+2)), express it as A/x + B/(x+2), leading to 5x+1 = A(x+2) + B x. Solving gives A = 1/2 and B = 9/2 by substituting x=0 and x=-2. Integrating each term yields (1/2) ln|x| + (9/2) ln|x+2| + C. A tempting distractor is choice E, which swaps signs, but this fails because it does not match the positive coefficients from solving. In general, for partial fractions with distinct linear factors, solve the system for coefficients accurately and verify by recombining or differentiating the antiderivative.

Question 8

1. $\frac{1}{5}\ln\left|\frac{x+5}{x-5}\right|+C$
2. $\frac{1}{5}\ln\left|\frac{x-5}{x+5}\right|+C$ (correct answer)
3. $\frac{1}{5}\ln|x-5|+\frac{1}{5}\ln|x+5|+C$
4. $\ln|x-5| - \ln|x+5|+C$
5. $\frac{2}{(x+5)(x-5)}+C$

Explanation: This problem requires the use of partial fraction decomposition to integrate a rational function. To evaluate $∫ \frac{2}{(x+5)(x-5)} \, dx$, decompose the integrand as $\frac{2}{(x+5)(x-5)} = \frac{A}{x+5} + \frac{B}{x-5}$. Solving for the coefficients gives $A = -\frac{1}{5}$ and $B = \frac{1}{5}$. Integrating term by term yields $- \frac{1}{5} \ln |x+5| + \frac{1}{5} \ln |x-5| + C$, which simplifies to $\frac{1}{5} \ln\left| \frac{x-5}{x+5} \right| + C$. A tempting distractor is choice A, $\frac{1}{5} \ln\left| \frac{x+5}{x-5} \right| + C$, but its derivative is $- \frac{2}{(x+5)(x-5)}$, the negative of the integrand. A transferable partial-fraction strategy is to multiply both sides by the common denominator and solve the resulting linear system for the coefficients.

Question 9

1. $\frac{3}{5}\ln|x-1|+\frac{7}{5}\ln|x+4|+C$ (correct answer)
2. $\frac{3}{5}\ln|x-1|-\frac{7}{5}\ln|x+4|+C$
3. $\frac{7}{5}\ln|x-1|+\frac{3}{5}\ln|x+4|+C$
4. $\frac{7}{5}\ln|x-1|-\frac{3}{5}\ln|x+4|+C$
5. $\ln|x-1|+\ln|x+4|+C$

Explanation: This problem uses partial fraction decomposition to integrate a rational function. We write $\frac{2x+1}{(x-1)(x+4)} = \frac{A}{x-1} + \frac{B}{x+4}$ and multiply through by $(x-1)(x+4)$ to get $2x+1 = A(x+4) + B(x-1)$. Setting $x=1$ gives $3 = 5A$, so $A = \frac{3}{5}$; setting $x=-4$ gives $-7 = -5B$, so $B = \frac{7}{5}$. Therefore, $\int \frac{2x+1}{(x-1)(x+4)}\,dx = \frac{3}{5}\ln|x-1| + \frac{7}{5}\ln|x+4| + C$. Choice E incorrectly assumes equal coefficients of 1 for both logarithms. Always solve for the specific constants A and B rather than guessing.

Question 10

1. $\frac{17}{4}\ln|t-3|+\frac{7}{4}\ln|t+1|+C$ (correct answer)
2. $\frac{17}{4}\ln|t-3|-\frac{7}{4}\ln|t+1|+C$
3. $\frac{7}{4}\ln|t-3|+\frac{17}{4}\ln|t+1|+C$
4. $\frac{7}{4}\ln|t-3|-\frac{17}{4}\ln|t+1|+C$
5. $6\ln|t-3|-\ln|t+1|+C$

Explanation: This integral requires partial fraction decomposition. We write $\frac{6t-1}{(t-3)(t+1)} = \frac{A}{t-3} + \frac{B}{t+1}$ and multiply by $(t-3)(t+1)$ to get $6t-1 = A(t+1) + B(t-3)$. Setting $t=3$ gives $17 = 4A$, so $A = \frac{17}{4}$; setting $t=-1$ yields $-7 = -4B$, so $B = \frac{7}{4}$. Therefore, $\int \frac{6t-1}{(t-3)(t+1)}\,dt = \frac{17}{4}\ln|t-3| + \frac{7}{4}\ln|t+1| + C$. Choice E incorrectly uses coefficients 6 and -1 without proper decomposition. The key is to clear denominators and evaluate at the roots of each factor.

Question 11

1. $\frac{1}{2}\ln\left|\frac{x-6}{x+2}\right|+C$ (correct answer)
2. $\frac{1}{2}\ln\left|\frac{x+2}{x-6}\right|+C$
3. $\frac{1}{2}\ln|x-6|+\frac{1}{2}\ln|x+2|+C$
4. $\ln|x-6| - \ln|x+2|+C$
5. $\frac{4}{(x+2)(x-6)}+C$

Explanation: This problem requires the use of partial fraction decomposition to integrate a rational function. To evaluate ∫ 4 / ((x+2)(x-6)) dx, decompose the integrand as 4 / ((x+2)(x-6)) = A/(x+2) + B/(x-6). Solving for the coefficients by substituting x = -2 and x = 6 gives A = -1/2 and B = 1/2. Integrating term by term yields (1/2) ln |x-6| - (1/2) ln |x+2| + C, which simplifies to (1/2) ln |(x-6)/(x+2)| + C. A tempting distractor is choice C, (1/2) ln |x-6| + (1/2) ln |x+2| + C, but this represents (1/2) ln |(x-6)(x+2)| + C, whose derivative is 4(x-2) / ((x+2)^2 (x-6)^2) rather than the original integrand. A transferable partial-fraction strategy is to multiply both sides by the common denominator and solve the resulting linear system for the coefficients.

Question 12

1. $\ln|x|+\ln|x+1|+C$ (correct answer)
2. $\ln\left|\frac{x}{x+1}\right|+C$
3. $2\ln|x|+\ln|x+1|+C$
4. $\ln|x|+2\ln|x+1|+C$
5. $\frac{2x+1}{x(x+1)}+C$

Explanation: This problem requires the use of partial fraction decomposition to integrate the rational function in a growth model. To decompose (2x+1)/(x(x+1)), express it as A/x + B/(x+1), leading to 2x+1 = A(x+1) + B x. Solving gives A = 1 and B = 1 by substituting x=0 and x=-1. Integrating each term yields ln|x| + ln|x+1| + C. A tempting distractor is choice B, which combines them incorrectly, but this fails because it does not match the coefficients from decomposition. In general, for partial fractions with distinct linear factors, solve the system for coefficients accurately and verify by recombining or differentiating the antiderivative.

Question 13

1. $\ln|x|+3\ln|x+1|+C$ (correct answer)
2. $3\ln|x|+\ln|x+1|+C$
3. $\ln|x| - 3\ln|x+1|+C$
4. $\ln\left|\frac{x}{x+1}\right|+C$
5. $\frac{4x+1}{x(x+1)}+C$

Explanation: This problem requires the use of partial fraction decomposition to integrate a rational function. To evaluate ∫ (4x+1) / (x(x+1)) dx, decompose the integrand as (4x+1) / (x(x+1)) = A/x + B/(x+1). Solving for the coefficients gives A = 1 and B = 3. Integrating term by term yields ln |x| + 3 ln |x+1| + C. A tempting distractor is choice B, 3 ln |x| + ln |x+1| + C, but its derivative is (4x + 3) / (x(x+1)), not matching 4x+1. A transferable partial-fraction strategy is to multiply both sides by the common denominator and solve the resulting linear system for the coefficients.

Question 14

1. $\ln|x|+\ln|x-3|+C$ (correct answer)
2. $\ln|x| - \ln|x-3|+C$
3. $2\ln|x| - \ln|x-3|+C$
4. $\ln|x| - 2\ln|x-3|+C$
5. $\frac{2x-3}{x(x-3)}+C$

Explanation: This problem requires the use of partial fraction decomposition to integrate a rational function. To evaluate ∫ (2x-3) / (x(x-3)) dx, decompose the integrand as (2x-3) / (x(x-3)) = A/x + B/(x-3). Solving for the coefficients gives A = 1 and B = 1. Integrating term by term yields ln |x| + ln |x-3| + C. A tempting distractor is choice B, ln |x| - ln |x-3| + C, but its derivative is (3) / (x(x-3)), not matching 2x-3 / (x(x-3)). A transferable partial-fraction strategy is to multiply both sides by the common denominator and solve the resulting linear system for the coefficients.

Question 15

1. $\ln|x|+2\ln|x+2|+C$ (correct answer)
2. $2\ln|x|+\ln|x+2|+C$
3. $\ln|x| - 2\ln|x+2|+C$
4. $2\ln|x+2| - \ln|x|+C$
5. $\frac{3x+2}{x(x+2)}+C$

Explanation: This problem requires the use of partial fraction decomposition to integrate a rational function. To evaluate ∫ (3x+2) / (x(x+2)) dx, decompose the integrand as (3x+2) / (x(x+2)) = A/x + B/(x+2). Solving for the coefficients gives A = 1 and B = 2. Integrating term by term yields ln |x| + 2 ln |x+2| + C. A tempting distractor is choice B, 2 ln |x| + ln |x+2| + C, but its derivative is (3x + 4) / (x(x+2)), not matching 3x+2. A transferable partial-fraction strategy is to multiply both sides by the common denominator and solve the resulting linear system for the coefficients.

Question 16

1. $\frac{1}{4}\ln\left|\frac{x-8}{x}\right|+C$ (correct answer)
2. $\frac{1}{4}\ln\left|\frac{x}{x-8}\right|+C$
3. $\frac{1}{4}\ln|x|+\frac{1}{4}\ln|x-8|+C$
4. $\ln|x| - \ln|x-8|+C$
5. $\frac{2}{x(x-8)}+C$

Explanation: This problem requires the use of partial fraction decomposition to integrate a rational function. To evaluate $∫ \frac{2}{x(x-8)} \, dx$, decompose the integrand as $\frac{2}{x(x-8)} = \frac{A}{x} + \frac{B}{x-8}$. Solving for the coefficients gives $A = -\frac{1}{4}$ and $B = \frac{1}{4}$. Integrating term by term yields $-\frac{1}{4} \ln |x| + \frac{1}{4} \ln |x-8| + C$, which simplifies to $\frac{1}{4} \ln\left| \frac{x-8}{x} \right| + C$. A tempting distractor is choice B, $\frac{1}{4} \ln\left| \frac{x}{x-8} \right| + C$, but its derivative is $-\frac{2}{x(x-8)}$, the negative of the integrand. A transferable partial-fraction strategy is to multiply both sides by the common denominator and solve the resulting linear system for the coefficients.

Question 17

1. $\frac{3}{5}\ln\left|\frac{x+4}{x+9}\right|+C$ (correct answer)
2. $\frac{3}{5}\ln\left|\frac{x+9}{x+4}\right|+C$
3. $\frac{3}{5}\ln|x+4|+\frac{3}{5}\ln|x+9|+C$
4. $\frac{3}{(x+4)(x+9)}+C$
5. $\frac{6}{5}\ln\left|\frac{x+4}{x+9}\right|+C$

Explanation: This problem requires the use of partial fraction decomposition to integrate a rational function. To evaluate $\int \frac{3}{(x+4)(x+9)} \, dx$, decompose the integrand as $\frac{3}{(x+4)(x+9)} = \frac{A}{x+4} + \frac{B}{x+9}$. Solving for the coefficients gives $A = \frac{3}{5}$ and $B = -\frac{3}{5}$. Integrating term by term yields $\frac{3}{5} \ln |x+4| - \frac{3}{5} \ln |x+9| + C$, which simplifies to $\frac{3}{5} \ln\left| \frac{x+4}{x+9} \right| + C$. A tempting distractor is choice B, $\frac{3}{5} \ln\left| \frac{x+9}{x+4} \right| + C$, but its derivative is $-\frac{3}{(x+4)(x+9)}$, the negative of the integrand. A transferable partial-fraction strategy is to multiply both sides by the common denominator and solve the resulting linear system for the coefficients.

Question 18

1. $\ln|x|+3\ln|x-1|+C$ (correct answer)
2. $3\ln|x|+\ln|x-1|+C$
3. $\ln|x| - 3\ln|x-1|+C$
4. $3\ln|x-1| - \ln|x|+C$
5. $\frac{4x-1}{x(x-1)}+C$

Explanation: This problem requires the use of partial fraction decomposition to integrate a rational function. To evaluate ∫ (4x-1) / (x(x-1)) dx, decompose the integrand as (4x-1) / (x(x-1)) = A/x + B/(x-1). Solving for the coefficients gives A = 1 and B = 3. Integrating term by term yields ln |x| + 3 ln |x-1| + C. A tempting distractor is choice B, 3 ln |x| + ln |x-1| + C, but its derivative is (4x + 1) / (x(x-1)), not matching 4x-1. A transferable partial-fraction strategy is to multiply both sides by the common denominator and solve the resulting linear system for the coefficients.

Question 19

1. $\ln|t-1|+C$
2. $\frac{5}{3}\ln|t-1|-\frac{2}{3}\ln|t+2|+C$ (correct answer)
3. $\frac{2}{3}\ln|t-1|+\frac{5}{3}\ln|t+2|+C$
4. $\frac{5}{3}\ln|t+2|-\frac{2}{3}\ln|t-1|+C$
5. $\frac{5}{3}\ln|t-1|+\frac{2}{3}\ln|t+2|+C$

Explanation: The skill used here is partial fraction decomposition to integrate the rational function. Factor the denominator as t² + t - 2 = (t - 1)(t + 2), and express (t + 4)/((t - 1)(t + 2)) as A/(t - 1) + B/(t + 2). Solving the equation t + 4 = A(t + 2) + B(t - 1) yields A = 5/3 and B = -2/3. Integrating term by term gives (5/3) ln|t - 1| - (2/3) ln|t + 2| + C. A tempting distractor like choice A is simply ln|t - 1| + C, which ignores the contribution from the other factor and the calculated coefficients. Always solve for the constants in partial fractions by substituting the roots or equating coefficients to ensure accurate integration of rational functions.

Question 20

1. $-\frac{8}{7}\ln|x-2|+\frac{15}{7}\ln|x+5|+C$ (correct answer)
2. $\frac{8}{7}\ln|x-2|+\frac{15}{7}\ln|x+5|+C$
3. $\frac{15}{7}\ln|x-2|-\frac{8}{7}\ln|x+5|+C$
4. $-\frac{15}{7}\ln|x-2|+\frac{8}{7}\ln|x+5|+C$
5. $\ln|x-2|+\ln|x+5|+C$

Explanation: This integral requires partial fraction decomposition. We write $\frac{x-10}{(x-2)(x+5)} = \frac{A}{x-2} + \frac{B}{x+5}$ and clear denominators to get $x-10 = A(x+5) + B(x-2)$. Setting $x=2$ gives $-8 = 7A$, so $A = -\frac{8}{7}$; setting $x=-5$ yields $-15 = -7B$, so $B = \frac{15}{7}$. Therefore, $\int \frac{x-10}{(x-2)(x+5)}\,dx = -\frac{8}{7}\ln|x-2| + \frac{15}{7}\ln|x+5| + C$. Choice E incorrectly assumes unit coefficients without decomposition. The key insight is that negative constants in partial fractions lead to negative coefficients in the integrated result.