Calculus 2 Quiz: Integration Long Division And Completing Square

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Integration Long Division And Completing SquareQuestion 1 of 20

To evaluate the integral $\int \frac{1}{ax^2+bx+c} dx$ where $b^2-4ac < 0$ , the technique of completing the square is used to transform the denominator into the form $a((x-h)^2+k^2)$ . What is the value of $h$ ?

h = -\frac{b}{2a}

h = \frac{b}{2a}

h = -\frac{b}{a}

h = \frac{b^2-4ac}{4a}

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Calculus 2 Quiz

Calculus 2 Quiz: Integration Long Division And Completing Square

Practice Integration Long Division And Completing Square in Calculus 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Integration Long Division And Completing Square, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 2.

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.

All questions

Question 1

$h = -\frac{b}{2a}$ (correct answer)
$h = \frac{b}{2a}$
$h = -\frac{b}{a}$
$h = \frac{b^2-4ac}{4a}$

Explanation: To complete the square for

ax^2+bx+c

, we first factor out

a

a\left(x^2 + \frac{b}{a}x\right) + c

Then, we complete the square for the expression in the parentheses. We take half of the coefficient of

x

, which is

\frac{b}{2a}

, and square it.

a\left(x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2\right) + c

a\left(x + \frac{b}{2a}\right)^2 - a\left(\frac{b}{2a}\right)^2 + c

The expression is of the form

a(x-h)^2 + \text{constant}

. Comparing

(x-h)^2

with

\left(x + \frac{b}{2a}\right)^2

, we see that

h = -\frac{b}{2a}

. Distractor B has a sign error. Distractor C misses the factor of 2. Distractor D is related to the vertex's y-coordinate, not the x-coordinate (h).

Question 2

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?

$k=4$
$k=5$
$k=6$
$k=7$ (correct answer)

Explanation: 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.

Question 3

The integral $\int \frac{x^4 + 2x^3 - x^2 + 3x - 2}{x^2 + x - 1} dx$ can be evaluated using polynomial long division. What is the degree of the polynomial part of the result before adding the integral of the proper rational function?

The polynomial part has degree 1 with remainder of degree 1
The polynomial part has degree 2 with remainder of degree 1 (correct answer)
The polynomial part has degree 2 with remainder of degree 0
The polynomial part has degree 3 with remainder of degree 1

Explanation: When dividing a polynomial of degree 4 by a polynomial of degree 2, the quotient will have degree

4 - 2 = 2

, and the remainder will have degree less than 2 (i.e., degree 1 or 0). Since we're dividing

x^4 + 2x^3 - x^2 + 3x - 2

x^2 + x - 1

, we expect a quotient of degree 2. Performing the division systematically: the quotient will be of the form

ax^2 + bx + c

and the remainder will be of the form

dx + e

. The remainder typically has degree 1 unless there's special cancellation. Therefore, the polynomial part has degree 2 and the remainder has degree 1. Choice A incorrectly gives degree 1 for the polynomial part. Choice C incorrectly states the remainder has degree 0. Choice D incorrectly gives degree 3, which is impossible since

4 - 2 = 2

Question 4

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?

$\int 2 dx + \int \frac{x + 3}{x^2 - 2x + 5} dx$
$\int 2 dx + \int \frac{-x + 7}{x^2 - 2x + 5} dx$
$\int 2 dx + \int \frac{x - 3}{x^2 - 2x + 5} dx$ (correct answer)
$\int 2 dx + \int \frac{-3x + 7}{x^2 - 2x + 5} dx$

Explanation: 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.

Question 5

To evaluate $\int \frac{x^2 + 6x + 13}{x^2 + 6x + 10} dx$ , which of the following is the most efficient first step?

Complete the square in both numerator and denominator, then use substitution $u = x + 3$
Perform polynomial long division to get $1 + \frac{3}{x^2 + 6x + 10}$ , then integrate each term separately (correct answer)
Use partial fraction decomposition after factoring the denominator completely over the reals
Apply integration by parts with $u = x^2 + 6x + 13$ and $dv = \frac{1}{x^2 + 6x + 10} dx$

Explanation: 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.

Question 6

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)?

$\frac{15u - 26}{u^2 + 4}$
$\frac{15u + 34}{u^2 + 4}$
$\frac{15u + 4}{u^2 + 4}$ (correct answer)
$\frac{15u + 64}{u^2 + 4}$

Explanation: 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

Question 7

Which of the following integrals requires polynomial long division as the necessary first step before other integration techniques can be applied?

$\int \frac{x^2+3x-5}{x^3-1} dx$
$\int \frac{1}{x^2+4x+8} dx$
$\int \frac{x^4 - 2x^2 + 1}{x^2+1} dx$ (correct answer)
$\int \frac{2x+1}{x^2+x+1} dx$

Explanation: 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.

Question 8

Evaluate the integral: $\int \frac{e^t}{e^{2t} - 4e^t + 8} dt$

$\frac{1}{2}\arctan\left(\frac{e^t-2}{2}\right) + C$ (correct answer)
$\ln(e^{2t} - 4e^t + 8) + C$
$\frac{1}{4}\arctan\left(\frac{e^t-2}{4}\right) + C$
$\frac{1}{2}\ln|e^{2t} - 4e^t + 8| + C$

Explanation: 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.

Question 9

Evaluate the integral: $\int \frac{1}{9x^2 - 6x + 5} dx$

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

Explanation: 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.

Question 10

Evaluate the integral: $\int \frac{1}{\sqrt{3-2x-x^2}} dx$

$\arcsin\left(\frac{x+1}{2}\right) + C$ (correct answer)
$\frac{1}{2}\arcsin\left(\frac{x+1}{2}\right) + C$
$\arcsin\left(\frac{x-1}{2}\right) + C$
$\ln|\sqrt{3-2x-x^2}| + C$

Explanation: The integral involves the square root of a quadratic, so we complete the square.

3-2x-x^2 = 3 - (x^2+2x) = 3 - (x^2+2x+1-1) = 3 - ((x+1)^2-1) = 4 - (x+1)^2

The integral becomes:

\int \frac{1}{\sqrt{4-(x+1)^2}} dx = \int \frac{1}{\sqrt{2^2-(x+1)^2}} dx

Let

u = x+1

, so

du = dx

. The integral is in the standard arcsin form

\int \frac{1}{\sqrt{a^2-u^2}}du = \arcsin(\frac{u}{a})

, with

a=2

\arcsin\left(\frac{u}{2}\right) + C = \arcsin\left(\frac{x+1}{2}\right) + C

Distractor B includes an incorrect coefficient. Distractor C has a sign error in completing the square. Distractor D uses the wrong integration rule entirely.

Question 11

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?

$x^2 - x + \frac{1}{2}\ln(x^2+9) - \arctan\left(\frac{x}{3}\right) + C$ (correct answer)
$x^2 - x + \ln(x^2+9) - \frac{1}{3}\arctan\left(\frac{x}{3}\right) + C$
$x^2 - x + \frac{1}{2}\ln(x^2+9) - 3\arctan(x) + C$
$x^2 - x + \frac{1}{2}\ln(x^2+9) - \frac{1}{3}\arctan\left(\frac{x}{3}\right) + C$

Explanation: 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

x^2-x

. The integral of

\frac{x}{x^2+9}

\frac{1}{2}\ln(x^2+9)

using a u-substitution

u=x^2+9

. The integral of

\frac{3}{x^2+9}

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.

Question 12

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 $a, h, k$ ?

$a=2, h=3, k=\sqrt{5/2}$ (correct answer)
$a=2, h=-3, k=5/2$
$a=1, h=3, k=\sqrt{5}$
$a=2, h=3, k=5/2$

Explanation: 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

Question 13

The integral of $\frac{1}{x^2-6x+c}$ results in an arctangent function. What can be concluded about the constant $c$ ?

$c = 9$
$c > 9$ (correct answer)
$c < 9$
$c$ can be any real number.

Explanation: 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.

Question 14

Evaluate the integral: $\int \frac{x-1}{x^2+2x+2} dx$

$\frac{1}{2}\ln(x^2+2x+2) + 2\arctan(x+1) + C$
$\frac{1}{2}\ln(x^2+2x+2) - 2\arctan(x+1) + C$ (correct answer)
$\ln(x^2+2x+2) - \arctan(x+1) + C$
$\frac{1}{2}\ln(x^2+2x+2) - \arctan(x+1) + C$

Explanation: When you encounter a rational function where the numerator's degree is less than the denominator's degree, you'll typically use partial fraction decomposition or break the integral into manageable parts. Here, since the denominator is quadratic and irreducible, the key is to split the numerator strategically. Start by rewriting the numerator

x-1

in terms that relate to the denominator

x^2+2x+2

. Notice that the derivative of

x^2+2x+2

2x+2 = 2(x+1)

. This means you can write:

x-1 = \frac{1}{2}(2x+2) - 2 = \frac{1}{2}(2x+2) - 2

So the integral becomes:

\int \frac{x-1}{x^2+2x+2} dx = \int \frac{\frac{1}{2}(2x+2) - 2}{x^2+2x+2} dx = \frac{1}{2}\int \frac{2x+2}{x^2+2x+2} dx - 2\int \frac{1}{x^2+2x+2} dx

The first integral is

\frac{1}{2}\ln(x^2+2x+2)

since you have

\frac{f'(x)}{f(x)}

. For the second, complete the square:

x^2+2x+2 = (x+1)^2+1

. This gives

\int \frac{1}{(x+1)^2+1} dx = \arctan(x+1)

. Therefore:

\frac{1}{2}\ln(x^2+2x+2) - 2\arctan(x+1) + C

Choice A has

+2\arctan(x+1)

instead of

-2\arctan(x+1)

, missing the negative sign from our decomposition. Choice C incorrectly has coefficient 1 instead of

\frac{1}{2}

for the logarithm and coefficient -1 instead of -2 for arctangent. Choice D has the correct logarithm term but wrong coefficient for arctangent (-1 instead of -2). Strategy tip: When integrating rational functions with irreducible quadratic denominators, always look for ways to create the derivative of the denominator in the numerator, then handle the remaining constant term separately.

Question 15

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?

0
1
2 (correct answer)
3

Explanation: This question tests the understanding of the polynomial long division algorithm. When dividing a polynomial

P(x)

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.

Question 16

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 $k$ ?

4
7
13 (correct answer)
25

Explanation: 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.

Question 17

Consider the integral $\int \frac{x^3 + ax}{x^2+1} dx$ . For which value of $a$ is the antiderivative $\frac{1}{2}x^2 + 2\ln(x^2+1) + C$ ?

$a = 1$
$a = 2$
$a = 3$
$a = 5$ (correct answer)

Explanation: 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

Question 18

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 $\int \frac{P(x)}{Q(x)} dx$ ?

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

Explanation: 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.

Question 19

The antiderivative of a function $f(x)$ is $F(x) = x - \ln(x^2+2x+2) + C$ . Which of the following is the function $f(x)$ ?

$\frac{x^2+x}{x^2+2x+2}$
$\frac{x^2+2x}{x^2+2x+2}$
$\frac{x^2+3x+2}{x^2+2x+2}$
$\frac{x^2}{x^2+2x+2}$ (correct answer)

Explanation: 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}

Question 20

Which substitution is the most effective first step to evaluate $\int \frac{\cos(\theta)}{4 - \sin^2(\theta)} d\theta$ ?

$u = \sin(\theta)$ , which transforms the integral into one requiring partial fractions. (correct answer)
$u = \sin(\theta)$ , which transforms the integral into an inverse tangent form.
$u = \cos(\theta)$ , which simplifies the expression to a polynomial.
$u = 4 - \sin^2(\theta)$ , which leads to a logarithmic integral.

Explanation: 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.