Integrating, Long Division, Completing the Square - AP Calculus BC

1. A rate is $r(x)=\dfrac{x^2+4x+7}{x+2}$. Find an antiderivative for $\int r(x)\,dx$.
   Answer: $\tfrac12 x^2+2x+3\ln|x+2|+C$ This integration problem requires polynomial long division as preparatory algebra since the numerator's degree equals the denominator's degree. Dividing $x^2 + 4x + 7$ by $x + 2$ yields quotient $x + 2$ with remainder 3, so $\frac{x^2+4x+7}{x+2} = x + 2 + \frac{3}{x+2}$. This algebraic manipulation is essential because it transforms an improper rational function into a polynomial plus a simple rational term, each of which has a standard antiderivative. Integrating gives $\int (x + 2 + \frac{3}{x+2})dx = \frac{1}{2}x^2 + 2x + 3\ln|x+2| + C$. Choice C mistakenly keeps the remainder as $\frac{3}{x+2}$ instead of integrating it to $3\ln|x+2|$, confusing the integrand with the integral. When faced with rational functions where the numerator's degree is not less than the denominator's, always divide first to create integrable pieces.
2. In a model for charge, compute $\int \frac{x^2+1}{x-2}\,dx$. Which antiderivative is correct?
   Answer: $\frac{x^2}{2}+2x+5\ln|x-2|+C$ This problem demonstrates the importance of polynomial long division as preparation for integration. Since the degree of the numerator ($x^2+1$) equals the degree of the denominator ($x-2$), we must perform long division first. Dividing $x^2+1$ by $x-2$ gives quotient $x+2$ with remainder $5$, so $\frac{x^2+1}{x-2} = x+2+\frac{5}{x-2}$. Integrating term by term yields $\int(x+2+\frac{5}{x-2})dx = \frac{x^2}{2}+2x+5\ln|x-2|+C$. Choice D incorrectly treats $\frac{5}{x-2}$ as if it were a power function rather than recognizing it integrates to a logarithm. The essential preparation strategy is to always perform polynomial division when the numerator's degree is greater than or equal to the denominator's degree.
3. A lens model uses $\int \dfrac{dx}{x^2-4x+13}$; which antiderivative is correct?
   Answer: $\dfrac13\arctan\!\left(\dfrac{x-2}{3}\right)+C$ 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.
4. A control system uses $\int \dfrac{x+1}{x^2+2x+5}\,dx$. Which antiderivative results after completing the square?
   Answer: $\tfrac12\ln(x^2+2x+5)+C$ This problem requires completing the square as the preparatory algebra for integration. The denominator x² + 2x + 5 becomes (x+1)² + 4, but since the numerator x+1 matches half the derivative, no further splitting is needed. This setup directly leads to a logarithmic antiderivative without an arctangent term. Omitting the square completion might mislead to incorrect forms. A tempting distractor like choice C fails by including an unnecessary arctangent, which applies only with a mismatched numerator. Always complete the square and check numerator-derivative alignment before integrating to determine the proper antiderivative structure.
5. A probability kernel uses $\int \dfrac{1}{x^2-4x+8}\,dx$. Which antiderivative is correct?
   Answer: $\dfrac12\arctan\!\left(\dfrac{x-2}{2}\right)+C$ The skill here is completing the square in the quadratic denominator to prepare for integration. This algebraic technique is essential for non-factorable quadratics, rewriting them as a squared term plus a constant for arctangent recognition. For x² - 4x + 8, it becomes (x - 2)² + 4, facilitating the integral as (1/2) arctan((x - 2)/2). Skipping this could result in incorrect logarithmic antiderivatives. A tempting distractor is choice A, omitting the 1/2 factor, but its derivative would be twice the integrand. Always use completing the square on quadratics prior to integration for correct standard form matching.
6. Compute $\int \frac{x^2+1}{x-2}\,dx$; which antiderivative matches the required long-division setup?
   Answer: $\frac12x^2+2x+5\ln|x-2|+C$ 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.
7. A lens model requires $\int \frac{4}{x^2+6x+13}\,dx$. Which antiderivative is correct?
   Answer: $2\arctan\left(\frac{x+3}{2}\right)+C$ This integral requires completing the square as the essential algebraic preparation. The denominator $x^2+6x+13$ must be rewritten to reveal its structure: $x^2+6x+13 = (x+3)^2-9+13 = (x+3)^2+4 = (x+3)^2+2^2$. This transforms our integral to $\int\frac{4}{(x+3)^2+2^2}dx$, which matches the arctangent form. Using the formula $\int\frac{1}{u^2+a^2}du = \frac{1}{a}\arctan(\frac{u}{a})+C$ with $u=x+3$ and $a=2$, we get $4\cdot\frac{1}{2}\arctan(\frac{x+3}{2})+C = 2\arctan(\frac{x+3}{2})+C$. Choice A incorrectly doubles this result to $4\arctan(\frac{x+3}{2})+C$, missing that the factor $\frac{1}{a}=\frac{1}{2}$ is already applied. Always complete the square carefully and apply the arctangent formula with proper coefficients.
8. Compute $\int \dfrac{1}{x^2+2x+5}\,dx$; which antiderivative matches completing the square?
   Answer: $\dfrac12\arctan\!\left(\dfrac{x+1}{2}\right)+C$ The skill here is completing the square in the quadratic denominator to prepare for integration. This step is vital for quadratics that don't factor, converting them to a completable form for arctangent integration. Completing the square on $x^2 + 2x + 5$ yields $(x + 1)^2 + 4$, leading to the integral $\frac{1}{2} \arctan\left(\frac{x + 1}{2}\right)$. Without it, improper forms like plain logarithms might be attempted. A tempting distractor is choice A, using $\arctan(x + 1)$ without scaling, failing as its derivative doesn't account for the correct coefficient. Always prepare quadratics by completing the square before integrating to ensure proper antiderivatives.
9. A signal has intensity proportional to $\int \dfrac{dx}{x^2+6x+13}$. Which antiderivative is correct?
   Answer: $\tfrac12\arctan\!\left(\dfrac{x+3}{2}\right)+C$ 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.
10. Evaluate $\int \dfrac{x^2+4x+8}{x^2+4x+5}\,dx$ by expressing the quadratic as $(x+2)^2+1$.
    Answer: $x+3\arctan(x+2)+C$ 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.
11. Find an antiderivative of $\dfrac{x^2+2x+2}{x+1}$ representing total distance traveled.
    Answer: $\dfrac{1}{2} x^2+x+\ln|x+1|+C$ The skill here is using polynomial long division to prepare a rational function for integration. This step is essential since the numerator's degree equals the denominator's, enabling a split into polynomial and fractional terms. Dividing $(x^2 + 2x + 2)/(x + 1)$ gives $x + 1 + \frac{1}{x + 1}$, simplifying the integration process. Without this, direct integration could be mishandled. A tempting distractor is choice B, replacing the logarithm with $\frac{1}{x + 1}$, but its derivative introduces a negative squared term that mismatches. Always employ long division for rational functions before integration to obtain correct results.
12. For accumulated cost, evaluate $\int \frac{x^2-2x+9}{x-1}\,dx$. Which antiderivative is correct?
    Answer: $\frac{x^2}{2}-x+8\ln|x-1|+C$ 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.
13. Compute an antiderivative of $\displaystyle \int \frac{1}{x^2-6x+13}\,dx$ using completing the square.​​
    Answer: $\tfrac12\arctan\!\left(\tfrac{x-3}{2}\right)+C$ This problem requires completing the square as preparatory algebra since the quadratic denominator has no real roots. Rewriting $x^2-6x+13 = (x-3)^2+4$, the integral becomes $\int \frac{1}{(x-3)^2+4}\,dx$. Using the arctangent formula with $u = x-3$ and $a = 2$, we get $\frac{1}{2}\arctan(\frac{x-3}{2})+C$. Choice A incorrectly omits the factor of $\frac{1}{2}$ that comes from the coefficient 4 in the completed square form. The key strategy: always complete the square for irreducible quadratics, then apply the standard arctangent integral formula $\int \frac{1}{u^2+a^2}\,du = \frac{1}{a}\arctan(\frac{u}{a})+C$.
14. In a fluid model, compute $\int \frac{x^2+4x+10}{x+2}\,dx$. Which antiderivative is correct?
    Answer: $\frac{x^2}{2}+2x+6\ln|x+2|+C$ This problem requires polynomial long division as the essential preparatory step for integration. The rational function $\frac{x^2+4x+10}{x+2}$ has a numerator of degree 2 and denominator of degree 1, so division is necessary. Performing long division: $x^2+4x+10$ divided by $x+2$ gives quotient $x+2$ with remainder $6$, yielding $\frac{x^2+4x+10}{x+2} = x+2+\frac{6}{x+2}$. Integrating each term gives $\int(x+2+\frac{6}{x+2})dx = \frac{x^2}{2}+2x+6\ln|x+2|+C$. Choice B incorrectly calculates the remainder as 10 instead of 6, likely from an arithmetic error during division. The key preparation principle is to always check degrees and perform polynomial division when the numerator's degree equals or exceeds the denominator's degree.
15. What is an antiderivative of $\displaystyle \int \frac{x^2-1}{x+3}\,dx$ after performing the necessary division step?​
    Answer: $\tfrac12x^2-3x+8\ln|x+3|+C$ 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.
16. For the velocity model $v(t)=\dfrac{t^2+1}{t+2}$, what is an antiderivative $\int \dfrac{t^2+1}{t+2}\,dt$?
    Answer: $\tfrac12 t^2-2t+5\ln|t+2|+C$ 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.
17. A particle's velocity is $v(t)=\dfrac{t^2+3t+5}{t+1}$; which antiderivative gives displacement $\int v(t)\,dt$?
    Answer: $\dfrac12 t^2+2t+3\ln|t+1|+C$ The key skill here is using polynomial long division as preparatory algebra for integrating rational functions where the numerator degree equals or exceeds the denominator degree. This algebraic step is necessary because it decomposes the integrand into a polynomial plus a proper fraction, making integration straightforward by separating integrable terms. Without long division, the integral remains complicated, as direct antiderivative rules do not apply to improper rational functions. Specifically, dividing t² + 3t + 5 by t + 1 yields t + 2 + 3/(t + 1), allowing integration to (1/2)t² + 2t + 3 ln|t + 1| + C. A tempting distractor like choice B fails because it uses ln|t| instead of ln|t + 1|, which would arise from an incorrect remainder term in division. Always perform preparatory algebra like long division before integrating to ensure the integrand is in a form amenable to basic techniques.
18. In a kinematics derivation, evaluate $\int \frac{x^2+2x+6}{x+3}\,dx$. Which antiderivative is correct?
    Answer: $\frac{x^2}{2}-x+9\ln|x+3|+C$ 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.
19. A particle's acceleration is $a(t)=\dfrac{t^2-4t+10}{t^2-4t+5}$. Find $\int a(t)\,dt$.
    Answer: $t+5\arctan(t-2)+C$ This problem requires completing the square in both numerator and denominator as preparatory algebra. The expression $t^2 - 4t + 5$ becomes $(t-2)^2 + 1$ after completing the square, essential for recognizing the arctangent integral pattern. Since the degrees are equal, we divide: $\frac{t^2-4t+10}{t^2-4t+5} = 1 + \frac{5}{t^2-4t+5} = 1 + \frac{5}{(t-2)^2+1}$. The remaining fraction is in standard arctangent form with $u = t-2$. Integrating gives $\int (1 + \frac{5}{(t-2)^2+1})dt = t + 5\arctan(t-2) + C$. Choice B incorrectly shows a negative coefficient on the arctangent term, possibly from a sign error in the algebraic manipulation. Always complete the square in quadratic expressions to transform them into recognizable integral forms.
20. To find total work, evaluate $\int \frac{x^2+3x+5}{x+1}\,dx$. Which antiderivative is correct?
    Answer: $\frac{x^2}{2}+2x+3\ln|x+1|+C$ 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.

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