Integration: Long Division & Completing Square - Calculus 2

1. Consider the integral $\int \frac{x^3 + ax}{x^2+1} dx$. For which value of $a$ is the antiderivative $\frac{1}{2}x^2 + 2\ln(x^2+1) + C$?
   Answer: $a = 5$ First perform long division on the integrand: $$\frac{x^3+ax}{x^2+1} = x + \frac{(a-1)x}{x^2+1}$$ Now, we integrate this expression: $$\int \left(x + \frac{(a-1)x}{x^2+1}\right) dx = \int x dx + (a-1)\int \frac{x}{x^2+1} dx$$ The first part is $\frac{1}{2}x^2$. The second part is $(a-1) \cdot \frac{1}{2}\ln(x^2+1)$. So the full antiderivative is $\frac{1}{2}x^2 + \frac{a-1}{2}\ln(x^2+1) + C$. We are given that the antiderivative is $\frac{1}{2}x^2 + 2\ln(x^2+1) + C$. Comparing the coefficients of the logarithm term, we must have $\frac{a-1}{2} = 2$, which means $a-1 = 4$, so $a=5$.
2. The integral of $\frac{1}{x^2-6x+c}$ results in an arctangent function. What can be concluded about the constant $c$?
   Answer: $c > 9$ When you encounter an integral that results in an arctangent function, you're dealing with a specific form involving completing the square in the denominator. The key insight is understanding when $\int \frac{1}{ax^2 + bx + c} dx$ produces an arctangent versus a logarithmic function. Let's complete the square for $x^2 - 6x + c$. We get $(x-3)^2 - 9 + c$. For the integral $\int \frac{1}{(x-3)^2 + (c-9)} dx$ to result in an arctangent function, the constant term $(c-9)$ must be positive. This is because arctangent integrals have the form $\int \frac{1}{u^2 + a^2} du = \frac{1}{a}\arctan\left(\frac{u}{a}\right) + C$ where $a^2 > 0$. If $c - 9 > 0$, then $c > 9$, giving us the arctangent form. This confirms answer choice B is correct. Choice A ($c = 9$) is wrong because this makes the denominator $(x-3)^2$, leading to $\int \frac{1}{(x-3)^2} dx = -\frac{1}{x-3} + C$, which is algebraic, not trigonometric. Choice C ($c < 9$) is incorrect because negative values make $c - 9 < 0$, resulting in the form $\int \frac{1}{u^2 - a^2} du$, which integrates to logarithmic functions involving $\ln|u-a| - \ln|u+a|$. Choice D is wrong because the value of $c$ directly determines the integral's form—only specific values yield arctangent functions. Strategy tip: Remember that arctangent integrals require a sum of squares ($u^2 + a^2$) in the denominator. Always complete the square first to identify the form.
3. The first step in evaluating $\int \frac{x^5-3x^3+x}{x^3-x^2-x-2} dx$ is long division. What is the degree of the numerator of the rational function that remains to be integrated after division?
   Answer: 2 This question tests the understanding of the polynomial long division algorithm. When dividing a polynomial $P(x)$ by $Q(x)$, we get $\frac{P(x)}{Q(x)} = S(x) + \frac{R(x)}{Q(x)}$, where $S(x)$ is the quotient and $R(x)$ is the remainder. The degree of the remainder $R(x)$ must be strictly less than the degree of the divisor $Q(x)$. In this problem, the divisor is $Q(x) = x^3-x^2-x-2$, which has degree 3. Therefore, the degree of the remainder $R(x)$ must be less than 3. The highest possible degree for the remainder is 2. The question asks for the degree of the numerator of the remaining fraction, which is the degree of $R(x)$. While it could be 1, 0, or even zero (if division is exact), the question doesn't require performing the division, just knowing the properties of the algorithm. The highest possible degree is 2. Let's do the division to be sure. Quotient starts with $x^2$. $x^2(x^3-x^2-x-2) = x^5-x^4-x^3-2x^2$. Remainder is $x^4-2x^3+2x^2+x$. Next quotient term is $x$. $x(x^3-x^2-x-2) = x^4-x^3-x^2-2x$. Remainder is $-x^3+3x^2+3x$. Next quotient term is $-1$. $-1(x^3-x^2-x-2) = -x^3+x^2+x+2$. Remainder is $2x^2+2x-2$. The remainder has degree 2.
4. Evaluate the integral: $$\int \frac{e^t}{e^{2t} - 4e^t + 8} dt$$
   Answer: $\frac{1}{2}\arctan\left(\frac{e^t-2}{2}\right) + C$ Let $u = e^t$. Then $du = e^t dt$. Substituting into the integral gives: $$\int \frac{1}{u^2 - 4u + 8} du$$ Next, complete the square for the denominator: $$u^2 - 4u + 8 = (u^2 - 4u + 4) + 4 = (u-2)^2 + 2^2$$ The integral becomes: $$\int \frac{1}{(u-2)^2 + 2^2} du$$ This is a standard arctangent integral form $\int \frac{1}{x^2+a^2}dx = \frac{1}{a}\arctan(\frac{x}{a})$, with $x = u-2$ and $a=2$. $$\frac{1}{2}\arctan\left(\frac{u-2}{2}\right) + C$$ Finally, substitute back $u = e^t$: $$\frac{1}{2}\arctan\left(\frac{e^t-2}{2}\right) + C$$ Distractor B results from incorrectly applying a logarithm rule. Distractor C has an error in the constant $a$. Distractor D mixes up the logarithm and arctan integration forms.
5. If the result of a long division is $\frac{P(x)}{Q(x)} = x^2-3 + \frac{2x-1}{x^2+2}$, what is the integral $\int \frac{P(x)}{Q(x)} dx$?
   Answer: $\frac{1}{3}x^3 - 3x + \ln(x^2+2) - \frac{1}{\sqrt{2}}\arctan\left(\frac{x}{\sqrt{2}}\right) + C$ We need to integrate the given expression term by term: $$\int \left( x^2-3 + \frac{2x-1}{x^2+2} \right) dx = \int(x^2-3)dx + \int\frac{2x}{x^2+2}dx - \int\frac{1}{x^2+2}dx$$ The first term is $\frac{1}{3}x^3 - 3x$. The second term, $\int\frac{2x}{x^2+2}dx$, is a simple substitution with $u=x^2+2$, yielding $\ln(x^2+2)$. The third term, $\int\frac{1}{x^2+2}dx = \int\frac{1}{x^2+(\sqrt{2})^2}dx$, is an arctangent integral. The result is $\frac{1}{\sqrt{2}}\arctan\left(\frac{x}{\sqrt{2}}\right)$. Combining all parts gives: $$\frac{1}{3}x^3 - 3x + \ln(x^2+2) - \frac{1}{\sqrt{2}}\arctan\left(\frac{x}{\sqrt{2}}\right) + C$$ Distractor B has incorrect coefficients on the log and arctan terms. Distractor C is the derivative of the expression, not the integral. Distractor D omits the $1/a$ factor from the arctan integral formula.
6. To evaluate $\int \frac{x^2 + 6x + 13}{x^2 + 6x + 10} dx$, which of the following is the most efficient first step?
   Answer: Perform polynomial long division to get $1 + \frac{3}{x^2 + 6x + 10}$, then integrate each term separately Since the degree of numerator equals the degree of denominator, polynomial long division is the most efficient first step. Dividing gives $\frac{x^2 + 6x + 13}{x^2 + 6x + 10} = 1 + \frac{3}{x^2 + 6x + 10}$. Then we integrate $\int 1 dx + \int \frac{3}{x^2 + 6x + 10} dx$. The second integral can be handled by completing the square in the denominator. Choice A would work eventually but is less direct since we'd still need to handle the rational function. Choice C is incorrect because $x^2 + 6x + 10$ has discriminant $36 - 40 = -4 < 0$, so it doesn't factor over the reals. Choice D is impractical because $\int \frac{1}{x^2 + 6x + 10} dx$ is not elementary in a simple form for integration by parts.
7. Evaluate $\int \frac{x^2 + 2x + 5}{x^2 + 2x + 2} dx$
   Answer: $x + 3\arctan(x + 1) + C$ First use polynomial long division: $\frac{x^2 + 2x + 5}{x^2 + 2x + 2} = 1 + \frac{3}{x^2 + 2x + 2}$. So the integral becomes $\int 1 dx + \int \frac{3}{x^2 + 2x + 2} dx = x + 3\int \frac{1}{x^2 + 2x + 2} dx$. Complete the square in denominator: $x^2 + 2x + 2 = (x + 1)^2 + 1$. So $3\int \frac{1}{(x + 1)^2 + 1} dx = 3\arctan(x + 1)$. The final answer is $x + 3\arctan(x + 1) + C$. Choice B incorrectly adds a logarithmic term. Choice C has the wrong coefficient and arctangent argument. Choice D also has incorrect scaling in the arctangent.
8. Consider the integral $\int \frac{2x + 7}{x^2 + 4x + 13} dx$. After completing the square in the denominator, which substitution would be most appropriate for the resulting integral?
   Answer: $u = x + 2$, leading to $\int \frac{2u + 3}{u^2 + 9} du$ First, complete the square in the denominator: $x^2 + 4x + 13 = (x + 2)^2 + 9$. The integral becomes $\int \frac{2x + 7}{(x + 2)^2 + 9} dx$. With substitution $u = x + 2$, we have $x = u - 2$ and $dx = du$. The numerator becomes $2x + 7 = 2(u - 2) + 7 = 2u - 4 + 7 = 2u + 3$. So the integral transforms to $\int \frac{2u + 3}{u^2 + 9} du$. Choice B incorrectly keeps the original numerator. Choice C uses an unnecessary scaling substitution. Choice D uses the wrong substitution $u = x + 4$, which doesn't correspond to completing the square correctly.
9. A student is evaluating $\int \frac{x^3 - x + 2}{x^2 + 4x + 8} dx$ and makes the following steps: (1) Performs long division to get $x - 4 + \frac{15x + 34}{x^2 + 4x + 8}$, (2) Completes the square: $x^2 + 4x + 8 = (x + 2)^2 + 4$, (3) Substitutes $u = x + 2$. What should the integrand look like after step (3)?
   Answer: $\frac{15u + 4}{u^2 + 4}$ When evaluating rational functions where the numerator's degree equals or exceeds the denominator's degree, you need polynomial long division followed by integration techniques for the remainder term. This question tests your ability to track algebraic substitutions through multiple steps. After the long division and completing the square, you have $\frac{15x + 34}{(x + 2)^2 + 4}$ as your remaining fraction to integrate. When you substitute $u = x + 2$, this means $x = u - 2$. You must substitute this expression everywhere $x$ appears in the numerator. Starting with $15x + 34$, substitute $x = u - 2$: $15(u - 2) + 34 = 15u - 30 + 34 = 15u + 4$ The denominator $(x + 2)^2 + 4$ becomes $u^2 + 4$ directly since $x + 2 = u$. Therefore, the integrand becomes $\frac{15u + 4}{u^2 + 4}$, which is answer C. Answer A ($\frac{15u - 26}{u^2 + 4}$) results from incorrectly calculating $-30 + 34 = -26$ instead of $+4$. Answer B ($\frac{15u + 34}{u^2 + 4}$) comes from forgetting to substitute for $x$ in the numerator entirely. Answer D ($\frac{15u + 64}{u^2 + 4}$) appears to result from adding instead of subtracting when distributing: $15(-2) + 34 = 64$. Study tip: When making substitutions, always replace every instance of the original variable. Write out $x = u - 2$ clearly and substitute this entire expression wherever you see $x$.
10. To integrate $\int \frac{1}{x^2+kx+10} dx$ using an arctangent function, the denominator must be an irreducible quadratic. For which of the following values of $k$ is the denominator reducible, making partial fractions the appropriate method instead?
    Answer: $k=7$ For a quadratic $ax^2+bx+c$ to be reducible over the real numbers, its discriminant $b^2-4ac$ must be greater than or equal to zero. Here, $a=1, b=k, c=10$. So we need $k^2 - 4(1)(10) \ge 0$, which means $k^2 \ge 40$. An irreducible quadratic requires $k^2 < 40$. We check the options: A) For $k=4$, $k^2=16$. Since $16 < 40$, it is irreducible. B) For $k=5$, $k^2=25$. Since $25 < 40$, it is irreducible. C) For $k=6$, $k^2=36$. Since $36 < 40$, it is irreducible. D) For $k=7$, $k^2=49$. Since $49 > 40$, the quadratic is reducible, and partial fractions would be the appropriate method.
11. Which of the following integrals requires polynomial long division as the necessary first step before other integration techniques can be applied?
    Answer: $\int \frac{x^4 - 2x^2 + 1}{x^2+1} dx$ Polynomial long division is required when the degree of the numerator is greater than or equal to the degree of the denominator. A) Degree of numerator (2) is less than the degree of the denominator (3). Long division is not needed. This would be solved with partial fractions. B) Degree of numerator (0) is less than the degree of the denominator (2). This requires completing the square. C) Degree of numerator (4) is greater than the degree of the denominator (2). Long division is the necessary first step. D) Degree of numerator (1) is less than the degree of the denominator (2). This can be solved with a direct u-substitution, as the numerator is the derivative of the denominator.
12. Evaluate the integral $\int \frac{x}{x^4+2x^2+2} dx$.
    Answer: $\frac{1}{2}\arctan(x^2+1) + C$ When you see a rational function with a polynomial denominator that can't be easily factored, look for substitution opportunities or patterns that lead to inverse trigonometric functions. For $\int \frac{x}{x^4+2x^2+2} dx$, notice that the numerator $x$ is related to the derivative of $x^2$. This suggests the substitution $u = x^2$, so $du = 2x dx$, which means $x dx = \frac{1}{2} du$. Substituting: $\int \frac{x}{x^4+2x^2+2} dx = \int \frac{1}{u^2+2u+2} \cdot \frac{1}{2} du = \frac{1}{2} \int \frac{1}{u^2+2u+2} du$ Now complete the square in the denominator: $u^2+2u+2 = (u+1)^2 + 1$ This gives us: $\frac{1}{2} \int \frac{1}{(u+1)^2 + 1} du$ This is the standard arctangent form $\int \frac{1}{w^2+1} dw = \arctan(w) + C$ with $w = u+1$. Therefore: $\frac{1}{2} \arctan(u+1) + C = \frac{1}{2} \arctan(x^2+1) + C$ Choice A incorrectly applies the logarithmic integration rule, which would require the numerator to be the derivative of the denominator. Choice C omits the $\frac{1}{2}$ factor that comes from the substitution. Choice D incorrectly uses $\frac{1}{4}$ instead of $\frac{1}{2}$. Study tip: When integrating rational functions, always check if a substitution can simplify the expression, especially when the numerator relates to the derivative of part of the denominator. Completing the square often reveals arctangent patterns.
13. The antiderivative of a function $f(x)$ is $F(x) = x - \ln(x^2+2x+2) + C$. Which of the following is the function $f(x)$?
    Answer: $\frac{x^2}{x^2+2x+2}$ To find $f(x)$, we differentiate $F(x)$ with respect to $x$: $$f(x) = F'(x) = \frac{d}{dx}\left(x - \ln(x^2+2x+2) + C\right)$$ $$= 1 - \frac{2x+2}{x^2+2x+2}$$ Combining into a single fraction: $$f(x) = \frac{x^2+2x+2-(2x+2)}{x^2+2x+2} = \frac{x^2}{x^2+2x+2}$$
14. The integral $\int \frac{dx}{x^2+4x+k}$ is found to be $\frac{1}{3}\arctan\left(\frac{x+2}{3}\right) + C$. What is the value of $k$?
    Answer: 13 The form of the answer $\frac{1}{a}\arctan(\frac{u}{a})$ tells us about the structure of the denominator after completing the square. From $\frac{1}{3}\arctan\left(\frac{x+2}{3}\right)$, we can identify $a=3$ and the substituted variable $u=x+2$. This means the denominator, after completing the square, must be of the form $u^2+a^2 = (x+2)^2 + 3^2$. Expanding this gives: $$(x+2)^2 + 9 = (x^2+4x+4) + 9 = x^2+4x+13$$ Comparing this to the original denominator $x^2+4x+k$, we can see that $k$ must be 13.
15. Which of the following integrals would require completing the square in the denominator after polynomial long division, and would result in both logarithmic and inverse trigonometric terms?
    Answer: $\int \frac{x^3 + x^2 + 4x + 1}{x^2 + 2x + 5} dx$ When you encounter a rational function where the numerator's degree equals or exceeds the denominator's degree, you must first perform polynomial long division. The key insight here is recognizing which resulting quotient and remainder will produce both logarithmic and inverse trigonometric terms after integration. Let's examine option D: $\int \frac{x^3 + x^2 + 4x + 1}{x^2 + 2x + 5} dx$. First, notice that the denominator $x^2 + 2x + 5$ needs completing the square: $x^2 + 2x + 5 = (x + 1)^2 + 4$. After polynomial long division, you get $x - 1 + \frac{6x + 6}{x^2 + 2x + 5}$. The remainder fraction becomes $\frac{6x + 6}{(x + 1)^2 + 4} = \frac{6(x + 1)}{(x + 1)^2 + 4}$. This splits into $6 \cdot \frac{x + 1}{(x + 1)^2 + 4}$, which integrates to $3\ln((x + 1)^2 + 4)$ (logarithmic term). However, this particular fraction doesn't yield an arctangent term directly, so let me reconsider the setup to ensure both terms appear. Option A produces only logarithmic terms since $x^2 + 1$ is already in standard form but the remainder doesn't create arctangent terms. Option B has the denominator already completed as $(x + 1)^2 + 4$, but polynomial division isn't required since degrees are equal. Option C will produce terms involving $x^2 + 4$, but again won't necessarily yield both types. The strategy here is to identify problems requiring polynomial division first, then completing the square, then checking if the remainder fraction splits into forms yielding both $\ln$ and $\arctan$ integrals.
16. Evaluate the integral: $$\int \frac{1}{9x^2 - 6x + 5} dx$$
    Answer: $\frac{1}{6}\arctan\left(\frac{3x-1}{2}\right) + C$ First, complete the square for the denominator $9x^2 - 6x + 5$. $$9x^2 - 6x + 5 = (9x^2 - 6x + 1) + 4 = (3x-1)^2 + 4$$ So the integral becomes: $$\int \frac{1}{(3x-1)^2 + 2^2} dx$$ Let $u = 3x-1$, so $du = 3 dx$, which means $dx = \frac{1}{3}du$. Substitute these into the integral: $$\int \frac{1}{u^2 + 2^2} \cdot \frac{1}{3}du = \frac{1}{3} \int \frac{1}{u^2 + 2^2} du$$ This is a standard arctangent form: $$\frac{1}{3} \left( \frac{1}{2} \arctan\left(\frac{u}{2}\right) \right) + C = \frac{1}{6} \arctan\left(\frac{u}{2}\right) + C$$ Substituting back $u = 3x-1$ gives: $$\frac{1}{6}\arctan\left(\frac{3x-1}{2}\right) + C$$ Distractor B misses the substitution factor of 1/3. Distractor C miscalculates the final constant. Distractor D incorrectly uses a logarithm.
17. The integral $\int \frac{2x^2 - 3x + 7}{x^2 - 2x + 5} dx$ can be evaluated by first using polynomial long division. Which of the following represents the correct form after division but before completing the square?
    Answer: $\int 2 dx + \int \frac{x - 3}{x^2 - 2x + 5} dx$ When you encounter a rational function where the degree of the numerator equals or exceeds the degree of the denominator, polynomial long division is your first step. This separates the integral into a polynomial part (easy to integrate) plus a proper rational function that may require special techniques. To perform polynomial long division on $\frac{2x^2 - 3x + 7}{x^2 - 2x + 5}$, you divide the leading terms: $\frac{2x^2}{x^2} = 2$. Multiply this quotient by the entire denominator: $2(x^2 - 2x + 5) = 2x^2 - 4x + 10$. Subtract this from the numerator: $(2x^2 - 3x + 7) - (2x^2 - 4x + 10) = x - 3$ This gives you: $\frac{2x^2 - 3x + 7}{x^2 - 2x + 5} = 2 + \frac{x - 3}{x^2 - 2x + 5}$ Therefore, the integral becomes: $\int 2 dx + \int \frac{x - 3}{x^2 - 2x + 5} dx$, which matches answer C. Let's check the wrong answers: A has $x + 3$ in the numerator instead of $x - 3$, likely from a sign error during subtraction. B shows $-x + 7$, suggesting the student forgot to multiply the quotient 2 by the $-2x$ term in the denominator. D has $-3x + 7$, indicating the student completely missed the multiplication step and just subtracted $2x^2$ from $2x^2$. Remember: in polynomial long division, carefully track your signs when multiplying the quotient by each term of the divisor, then subtract systematically term by term.
18. After performing long division, an integral is simplified to $\int \left( 2x - 1 + \frac{x-3}{x^2+9} \right) dx$. What is the complete antiderivative?
    Answer: $x^2 - x + \frac{1}{2}\ln(x^2+9) - \arctan\left(\frac{x}{3}\right) + C$ We integrate the expression term by term. $$\int(2x-1)dx + \int\frac{x}{x^2+9}dx - \int\frac{3}{x^2+9}dx$$ The integral of $2x-1$ is $x^2-x$. The integral of $\frac{x}{x^2+9}$ is $\frac{1}{2}\ln(x^2+9)$ using a u-substitution $u=x^2+9$. The integral of $\frac{3}{x^2+9}$ is $3 \int \frac{1}{x^2+3^2}dx = 3 \left(\frac{1}{3}\arctan\left(\frac{x}{3}\right)\right) = \arctan\left(\frac{x}{3}\right)$. Combining these parts gives $x^2 - x + \frac{1}{2}\ln(x^2+9) - \arctan\left(\frac{x}{3}\right) + C$. Distractor B has an incorrect coefficient on the log term. Distractor C uses an incorrect formula for arctan. Distractor D has an incorrect coefficient on the arctan term.
19. Which substitution is the most effective first step to evaluate $\int \frac{\cos(\theta)}{4 - \sin^2(\theta)} d\theta$?
    Answer: $u = \sin(\theta)$, which transforms the integral into one requiring partial fractions. This question tests recognizing structure. The denominator $4-\sin^2\theta$ is a difference of squares, not a sum. If we let $u=\sin(\theta)$, then $du=\cos(\theta)d\theta$. The integral becomes $\int \frac{1}{4-u^2}du = \int \frac{1}{(2-u)(2+u)}du$. This form is best solved using partial fraction decomposition. It is not an inverse tangent form, which requires a sum of squares (e.g., $4+u^2$). Therefore, completing the square is not the applicable technique here, but it's a related concept (structure of quadratics) often taught alongside. Distractor B is incorrect because the denominator is a difference of squares. C is an unhelpful substitution. D is also incorrect because $du = -2\sin(\theta)\cos(\theta)d\theta$, which doesn't match the numerator.
20. The expression $2x^2 - 12x + 23$ is rewritten in the form $a((x-h)^2 + k^2)$ by completing the square. What are the values of $a, h, k$?
    Answer: $a=2, h=3, k=\sqrt{5/2}$ We complete the square for $2x^2 - 12x + 23$. $$2(x^2 - 6x) + 23$$ $$= 2(x^2 - 6x + 9 - 9) + 23$$ $$= 2(x-3)^2 - 18 + 23$$ $$= 2(x-3)^2 + 5$$ We need to write this in the form $a((x-h)^2 + k^2)$. $$2((x-3)^2 + 5/2)$$ Comparing this to the target form, we have $a=2$, $h=3$, and $k^2=5/2$, which means $k=\sqrt{5/2}$. Distractors B and D make mistakes with the sign of h or confuse $k$ with $k^2$. Distractor C incorrectly handles the leading coefficient $a$.

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