AP Calculus BC Quiz: Integrating Long Division Completing The Square

What this quiz covers

This quiz focuses on Integrating Long Division Completing The Square, 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. $\dfrac12\ln(x^2-4x+13)+C$
2. $\dfrac13\arctan\!\left(\dfrac{x-2}{3}\right)+C$ (correct answer)
3. $\arctan\!\left(\dfrac{x-2}{3}\right)+C$
4. $\ln|x^2-4x+13|+C$
5. $\dfrac{1}{x^2-4x+13}+C$

Explanation: The key skill here is completing the square as preparatory algebra for integrating rational functions with quadratic denominators. This algebraic manipulation is crucial to transform the denominator into (x - h)^2 + k, revealing an arctangent form for the integral. Neglecting it prevents matching the integrand to known antiderivative patterns. The result is (1/3) arctan((x - 2)/3) + C after adjusting the coefficient. A tempting distractor like choice C fails because it lacks the 1/3 scalar, incorrectly scaling the arctangent. Always perform preparatory algebra like completing the square before integrating to ensure the integrand is in a form amenable to basic techniques.

Question 2

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

Explanation: This integral requires completing the square as the crucial preparatory step. The denominator $x^2-4x+8$ is an irreducible quadratic that must be rewritten in the form $(x-a)^2+b^2$. Completing the square: $x^2-4x+8 = (x-2)^2-4+8 = (x-2)^2+4 = (x-2)^2+2^2$. This transforms our integral to $\int\frac{3}{(x-2)^2+2^2}dx$, which matches the arctangent form $\int\frac{1}{u^2+a^2}du = \frac{1}{a}\arctan(\frac{u}{a})+C$. Applying this with $u=x-2$ and $a=2$ gives $3\cdot\frac{1}{2}\arctan(\frac{x-2}{2})+C = \frac{3}{2}\arctan(\frac{x-2}{2})+C$. Choice C incorrectly assumes this leads to a logarithm rather than arctangent. Always complete the square for irreducible quadratics in the denominator to reveal the appropriate integration form.

Question 3

1. $\tfrac12(t+2)^2-4\ln|t+2|+C$
2. $\tfrac12 t^2-2t+\ln|t+2|+C$
3. $\tfrac12 t^2-2t+5\ln|t+2|+C$ (correct answer)
4. $\tfrac12 t^2+2t+5\ln|t+2|+C$
5. $\tfrac12 t^2-2t+\dfrac{5}{t+2}+C$

Explanation: This problem requires algebraic preparation through polynomial long division before integrating the rational function. Since the degree of the numerator (2) equals the degree of the denominator (1), we must divide $t^2 + 1$ by $t + 2$ to get $t - 2$ with remainder 5, giving us $\frac{t^2+1}{t+2} = t - 2 + \frac{5}{t+2}$. This decomposition is necessary because we cannot directly integrate the original rational expression—we need it in a form where each piece has a known antiderivative. Integrating term by term yields $\int (t - 2 + \frac{5}{t+2})dt = \frac{1}{2}t^2 - 2t + 5\ln|t+2| + C$. Choice E incorrectly treats the remainder as $\frac{5}{t+2}$ in the final answer rather than integrating it to get $5\ln|t+2|$. Always perform polynomial division when the numerator's degree is greater than or equal to the denominator's degree before attempting integration.

Question 4

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

Explanation: The skill here is using polynomial long division to prepare a rational function for integration. This algebraic manipulation is crucial when the numerator degree is higher than or equal to the denominator's, yielding a polynomial plus a proper rational term. For (2x² + 3x - 1)/(x + 1), division produces 2x + 1 - 2/(x + 1), separating into integrable components. Omitting this step could lead to flawed integration attempts. A tempting distractor is choice B, which replaces the logarithm with -2/(x + 1), but its derivative includes an incorrect squared term in the denominator. Always simplify with long division before integrating rational functions to ensure accuracy.

Question 5

1. $\arctan\!\left(\dfrac{x+3}{2}\right)+C$
2. $\dfrac12\arctan\!\left(\dfrac{x+3}{2}\right)+C$ (correct answer)
3. $\dfrac{1}{x^2+6x+13}+C$
4. $\ln(x^2+6x+13)+C$
5. $\dfrac12\ln(x^2+6x+13)+C$

Explanation: The skill here is completing the square in the quadratic denominator to prepare for integration. This step is necessary for irreducible quadratics, transforming the expression into a form suitable for the arctangent integral formula. Completing the square on x² + 6x + 13 gives (x + 3)² + 4, allowing a substitution that matches ∫ du/(u² + a²) with a = 2. Without this preparation, one might erroneously apply logarithmic integration rules. A tempting distractor is choice A, which lacks the 1/2 coefficient, failing because its derivative is twice the original integrand. Always complete the square for quadratic denominators before integrating to align with standard forms.

Question 6

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

Explanation: This integral requires completing the square as the essential preparatory algebra step. The denominator $x^2+4x+8$ must be rewritten in the form $(x-h)^2+k^2$ to enable integration using the arctangent formula. Completing the square: $x^2+4x+8 = (x+2)^2+4 = (x+2)^2+2^2$. This algebraic preparation is crucial because it transforms the integral into the standard form $\int\frac{1}{u^2+a^2}du = \frac{1}{a}\arctan(\frac{u}{a})+C$. With $u=x+2$ and $a=2$, we get $\int\frac{1}{(x+2)^2+4}dx = \frac{1}{2}\arctan(\frac{x+2}{2})+C$. The error of choosing $\ln|x^2+4x+8|+C$ would apply the wrong integration formula for this quadratic form. When the denominator is an irreducible quadratic (discriminant negative), always complete the square to reveal the arctangent structure.

Question 7

1. $\frac{1}{2}\arctan\!\left(\frac{x+4}{2}\right)+C$ (correct answer)
2. $\frac{1}{4}\arctan\!\left(\frac{x+4}{2}\right)+C$
3. $\ln|x^2+8x+20|+C$
4. $\arctan\!\left(\frac{x+4}{2}\right)+C$
5. $\arctan(x^2+8x+20)+C$

Explanation: This integral demonstrates the necessity of completing the square as preparatory algebra before integration. We must rewrite $x^2+8x+20$ in standard form to apply the arctangent formula. Completing the square: $x^2+8x+20 = (x+4)^2+4 = (x+4)^2+2^2$. This algebraic preparation transforms the integral into the recognizable form $\int\frac{1}{u^2+a^2}du = \frac{1}{a}\arctan(\frac{u}{a})+C$. With $u=x+4$ and $a=2$, we get $\int\frac{1}{(x+4)^2+4}dx = \frac{1}{2}\arctan(\frac{x+4}{2})+C$. The incorrect choice $\ln|x^2+8x+20|+C$ would result from confusing this with integrals having linear denominators. When integrating rational functions with irreducible quadratic denominators, always complete the square first to expose the underlying arctangent structure.

Question 8

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

Explanation: This problem requires algebraic preparation before integration can proceed. The rational function $\frac{x^2+3x+5}{x+1}$ has a numerator of higher degree than the denominator, so polynomial long division must be performed first. Dividing $x^2+3x+5$ by $x+1$ yields quotient $x+2$ with remainder $3$, giving us $\frac{x^2+3x+5}{x+1} = x+2+\frac{3}{x+1}$. Now we can integrate term by term: $\int(x+2+\frac{3}{x+1})dx = \frac{x^2}{2}+2x+3\ln|x+1|+C$. Choice D incorrectly attempts to use logarithm properties without first simplifying the rational expression. The key strategy is to always check if the degree of the numerator is greater than or equal to the denominator's degree—if so, perform polynomial division before integrating.

Question 9

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

Explanation: This problem demonstrates polynomial long division as the crucial preparatory step for integration. The rational function $\frac{x^2+2x+6}{x+3}$ has numerator degree 2 and denominator degree 1, necessitating division. Performing the division: $x^2+2x+6$ divided by $x+3$ yields quotient $x-1$ with remainder $9$, so $\frac{x^2+2x+6}{x+3} = x-1+\frac{9}{x+3}$. Integrating each term gives $\int(x-1+\frac{9}{x+3})dx = \frac{x^2}{2}-x+9\ln|x+3|+C$. Choice C incorrectly identifies the remainder as 6 instead of 9, likely from a calculation error during polynomial division. The key preparation principle is to systematically perform polynomial division whenever the numerator's degree equals or exceeds the denominator's degree.

Question 10

1. $\tfrac12 x^2+2x+3\ln|x-2|+C$ (correct answer)
2. $\tfrac12 x^2+2x-3\ln|x-2|+C$
3. $\tfrac12 x^2-2x+3\ln|x-2|+C$
4. $\tfrac12 x^2+2x+\dfrac{3}{x-2}+C$
5. $\tfrac12 (x-2)^2+3\ln|x-2|+C$

Explanation: This integration problem requires polynomial long division as preparatory algebra since the numerator's degree equals the denominator's degree. Dividing $x^2 - 1$ by $x - 2$ gives quotient $x + 2$ with remainder 3, so $\frac{x^2-1}{x-2} = x + 2 + \frac{3}{x-2}$. This division step is crucial because it transforms the improper rational function into a polynomial plus a simple fraction, each having standard antiderivatives. Integrating term by term yields $\int (x + 2 + \frac{3}{x-2})dx = \frac{1}{2}x^2 + 2x + 3\ln|x-2| + C$. Choice D incorrectly leaves the remainder as $\frac{3}{x-2}$ instead of integrating it to $3\ln|x-2|$, confusing the integrand with its integral. When integrating rational functions, always check if long division is needed by comparing the degrees of numerator and denominator.

Question 11

1. $\frac{x^2}{2}-x+8\ln|x-1|+C$ (correct answer)
2. $\frac{x^2}{2}-x+9\ln|x-1|+C$
3. $\frac{x^2}{2}-2x+8\ln|x-1|+C$
4. $\frac{x^2}{2}-x+\frac{8}{x-1}+C$
5. $\ln\left|\frac{x^2-2x+9}{x-1}\right|+C$

Explanation: This problem illustrates the necessity of polynomial long division as preparation for integration. With $\frac{x^2-2x+9}{x-1}$, the numerator and denominator have degrees 2 and 1 respectively, requiring division. Dividing $x^2-2x+9$ by $x-1$: the quotient is $x-1$ with remainder $8$, giving us $\frac{x^2-2x+9}{x-1} = x-1+\frac{8}{x-1}$. Integrating term by term yields $\int(x-1+\frac{8}{x-1})dx = \frac{x^2}{2}-x+8\ln|x-1|+C$. Choice B incorrectly computes the remainder as 9 instead of 8, likely from an error in the division process. The fundamental preparation strategy remains: always perform polynomial division when the numerator's degree is at least as large as the denominator's degree.

Question 12

1. $\arctan\!\left(\dfrac{x+3}{2}\right)+C$
2. $\tfrac12\arctan\!\left(\dfrac{x+3}{2}\right)+C$ (correct answer)
3. $\tfrac12\ln(x^2+6x+13)+C$
4. $\dfrac{1}{x^2+6x+13}+C$
5. $2\arctan\!\left(\dfrac{x+3}{2}\right)+C$

Explanation: This problem requires completing the square as the preparatory algebra for integration. The quadratic denominator x² + 6x + 13 has a negative discriminant, so it doesn't factor over the reals and needs rewriting as (x+3)² + 4. This form matches the standard integral for 1/(u² + a²), yielding an arctangent antiderivative scaled by 1/a. Without completing the square, the integral cannot be matched to this form. A tempting distractor like choice A fails because it omits the 1/2 coefficient needed for the a=2 scaling. Always complete the square for irreducible quadratic denominators before integrating to access arctangent or logarithmic forms.

Question 13

1. $\tfrac12\ln(x^2+2x+5)+C$ (correct answer)
2. $\ln(x^2+2x+5)+C$
3. $\arctan\!\left(\frac{x+1}{2}\right)+C$
4. $\tfrac12\ln|x+1|+C$
5. $\tfrac12\ln(x^2+2x+5)+\arctan\!\left(\frac{x+1}{2}\right)+C$

Explanation: This integral demonstrates completing the square as essential preparatory algebra for integration. The denominator $x^2+2x+5 = (x+1)^2+4$ has no real roots, requiring this technique. With $u = x+1$, the integral becomes $\int \frac{u}{u^2+4}\,du$, which is the derivative form yielding $\frac{1}{2}\ln(u^2+4)+C = \frac{1}{2}\ln(x^2+2x+5)+C$. Choice C incorrectly suggests only an arctangent result, which would arise from $\int \frac{1}{(x+1)^2+4}\,dx$, not our numerator $x+1$. The fundamental principle: when denominators are irreducible quadratics, complete the square to reveal standard forms.

Question 14

1. $x+3\ln(x^2-6x+10)+4\arctan(x-3)+C$
2. $x+3\ln(x^2-6x+10)-4\arctan(x-3)+C$ (correct answer)
3. $x+3\ln|x-3|-4\arctan(x-3)+C$
4. $\ln(x^2+1)+C$
5. $x+3\ln(x^2-6x+10)-4\arctan\!\left(\frac{x}{\sqrt{10}}\right)+C$

Explanation: This integral demands polynomial long division and completing the square as preparatory algebra. Since both polynomials have degree 2, we divide: $\frac{x^2+1}{x^2-6x+10} = 1 + \frac{6x-9}{x^2-6x+10}$. The denominator $x^2-6x+10$ completes to $(x-3)^2+1$, preparing us for logarithmic and arctangent integration. We decompose the remaining fraction and integrate to get $x + 3\ln(x^2-6x+10) - 4\arctan(x-3) + C$, where the negative sign on the arctangent comes from the algebraic manipulation. Choice E incorrectly uses $\arctan\left(\frac{x}{\sqrt{10}}\right)$, failing to account for the completed square form $(x-3)^2+1$. Always complete the square in the denominator to reveal the correct argument for the arctangent function.

Question 15

1. $\tfrac12x^2+x+\ln|x+1|+C$ (correct answer)
2. $\tfrac12x^2+x+C$
3. $\ln|x^2+2x+2|+C$
4. $\tfrac12x^2+x+\arctan(x+1)+C$
5. $\tfrac12x^2+\ln|x+1|+C$

Explanation: This integral requires long division as preparatory algebra since the numerator and denominator have equal degrees. Dividing $(x^2+2x+2)÷(x+1)$ produces quotient $x+1$ with remainder $1$, transforming the integrand to $x+1+\frac{1}{x+1}$. Integration yields $\frac{1}{2}x^2+x+\ln|x+1|+C$. Choice D incorrectly introduces arctangent, which would only appear if we had a quadratic denominator like $(x+1)^2+1$ after completing the square. The essential strategy: recognize improper rational functions by comparing degrees, then use long division to create integrable pieces.

Question 16

1. $x+3\arctan(x-1)+C$ (correct answer)
2. $x+\tfrac32\arctan(x-1)+C$
3. $x-3\arctan(x-1)+C$
4. $\tfrac12\ln\big((x-1)^2+1\big)+3\arctan(x-1)+C$
5. $x+\tfrac32\ln\big((x-1)^2+1\big)+C$

Explanation: This integration problem requires completing the square in the denominator as preparatory algebra. The denominator $x^2 - 2x + 2$ can be rewritten as $(x-1)^2 + 1$, which is essential for recognizing arctangent integral patterns. Since the numerator and denominator have equal degrees, we first divide: $\frac{x^2-2x+5}{x^2-2x+2} = 1 + \frac{3}{x^2-2x+2} = 1 + \frac{3}{(x-1)^2+1}$. The fraction $\frac{3}{(x-1)^2+1}$ is now in standard arctangent form with $u = x-1$. Integrating gives $\int (1 + \frac{3}{(x-1)^2+1})dx = x + 3\arctan(x-1) + C$. Choice C incorrectly has a negative coefficient on the arctangent, perhaps from a sign error in the division process. When you see a quadratic denominator, always check if completing the square reveals a sum of squares form suitable for arctangent integration.

Question 17

1. $x+3\arctan(x+2)+C$ (correct answer)
2. $x-3\arctan(x+2)+C$
3. $\tfrac12\ln\big((x+2)^2+1\big)+3\arctan(x+2)+C$
4. $x+\tfrac12\ln\big((x+2)^2+1\big)+C$
5. $x+\arctan(x+2)+C$

Explanation: This integration requires completing the square in the denominator as essential preparatory algebra. The denominator $x^2 + 4x + 5$ becomes $(x+2)^2 + 1$ after completing the square, which allows us to use the arctangent integral formula. Since numerator and denominator have the same degree, we first divide: $\frac{x^2+4x+8}{x^2+4x+5} = 1 + \frac{3}{x^2+4x+5} = 1 + \frac{3}{(x+2)^2+1}$. The fraction $\frac{3}{(x+2)^2+1}$ is in standard arctangent form with substitution $u = x+2$. Integrating yields $\int (1 + \frac{3}{(x+2)^2+1})dx = x + 3\arctan(x+2) + C$. Choice B incorrectly has a negative coefficient on arctangent, likely from a computational error. When faced with quadratic denominators, always complete the square to reveal whether you have an arctangent or logarithm integral.

Question 18

1. $\frac12x^2+2x+5\ln|x-2|+C$ (correct answer)
2. $\frac12x^2+2x+\ln|x-2|+C$
3. $\frac12(x-2)^2+5\ln|x-2|+C$
4. $\ln|x^2+1|+C$
5. $\frac12x^2+2x+5\ln|x-2|+C$

Explanation: This integral requires polynomial long division as the essential preparatory algebra. Dividing $x^2+1$ by $x-2$ gives us $x+2$ with remainder $5$, so $\frac{x^2+1}{x-2} = x+2+\frac{5}{x-2}$. This step is necessary because the degree of the numerator (2) exceeds the degree of the denominator (1), preventing direct integration. Now we can integrate term by term: $\int(x+2+\frac{5}{x-2})dx = \frac{x^2}{2}+2x+5\ln|x-2|+C$. Choice D's suggestion of $\ln|x^2+1|+C$ incorrectly assumes the numerator is the derivative of some related expression, which it isn't. Always perform polynomial division when the numerator's degree is at least as large as the denominator's, then integrate the resulting polynomial and simple fraction separately.

Question 19

1. $\arctan\!\left(\dfrac{x-2}{2}\right)+C$
2. $\tfrac12\arctan\!\left(\dfrac{x-2}{2}\right)+C$ (correct answer)
3. $\tfrac12\ln(x^2-4x+8)+C$
4. $\dfrac{1}{x^2-4x+8}+C$
5. $2\arctan\!\left(\dfrac{x-2}{2}\right)+C$

Explanation: This problem requires completing the square as the preparatory algebra for integration. The denominator $x^2 - 4x + 8$ completes to $(x-2)^2 + 4$, revealing the sum-of-squares structure. This transformation is crucial for recognizing the integral as a scaled arctangent function. Ignoring this step prevents the substitution that simplifies the integral. A tempting distractor like choice C fails as it incorrectly suggests a logarithmic form, which applies only to differences of squares. Always complete the square in quadratic denominators prior to integration to identify the appropriate antiderivative type.

Question 20

1. $\tfrac12x^2-3x+8\ln|x+3|+C$ (correct answer)
2. $\tfrac12x^2-3x+\ln|x+3|+C$
3. $\tfrac12x^2+3x+8\ln|x+3|+C$
4. $\ln|x^2-1|+C$
5. $\tfrac13x^3-x+C$

Explanation: This problem illustrates the critical role of polynomial long division as preparatory algebra before integration. With numerator degree 2 and denominator degree 1, we must divide $(x^2-1)÷(x+3)$ first, yielding quotient $x-3$ with remainder $8$. This transforms the integrand to $x-3+\frac{8}{x+3}$, which integrates to $\frac{1}{2}x^2-3x+8\ln|x+3|+C$. Choice D's logarithm of the original numerator reveals a common misconception—this approach only works when the numerator is the derivative of the denominator. Always check degrees first: if the rational function is improper, division is the essential first step.