Calculus 1 Quiz: Integration Long Division And Completing Square

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Integration Long Division And Completing SquareQuestion 1 of 20

Which of the following is equivalent to $\int \frac{2x - 1}{x^2 - 6x + 13} dx$ ?

\ln(x^2 - 6x + 13) + \frac{5}{2}\arctan\left(\frac{x-3}{2}\right) + C

\ln(x^2 - 6x + 13) + 5\arctan\left(\frac{x-3}{2}\right) + C

\frac{1}{2}\ln(x^2 - 6x + 13) + \frac{5}{2}\arctan\left(\frac{x-3}{2}\right) + C

\ln(x^2 - 6x + 13) - \frac{7}{2}\arctan\left(\frac{x-3}{2}\right) + C

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Calculus 1 Quiz

Calculus 1 Quiz: Integration Long Division And Completing Square

Practice Integration Long Division And Completing Square in Calculus 1 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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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 1.

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Question 1

Which of the following is equivalent to $\int \frac{2x - 1}{x^2 - 6x + 13} dx$ ?

$\ln(x^2 - 6x + 13) + \frac{5}{2}\arctan\left(\frac{x-3}{2}\right) + C$ (correct answer)
$\ln(x^2 - 6x + 13) + 5\arctan\left(\frac{x-3}{2}\right) + C$
$\frac{1}{2}\ln(x^2 - 6x + 13) + \frac{5}{2}\arctan\left(\frac{x-3}{2}\right) + C$
$\ln(x^2 - 6x + 13) - \frac{7}{2}\arctan\left(\frac{x-3}{2}\right) + C$

Explanation: The denominator

x^2 - 6x + 13

is an irreducible quadratic. First, complete the square:

x^2 - 6x + 13 = (x^2 - 6x + 9) + 4 = (x-3)^2 + 2^2

. The derivative of the original denominator is

2x-6

. We manipulate the numerator to contain this term:

2x - 1 = (2x - 6) + 5

. Now, split the integral:

\int \frac{(2x - 6) + 5}{x^2 - 6x + 13} dx = \int \frac{2x - 6}{x^2 - 6x + 13} dx + \int \frac{5}{(x-3)^2 + 4} dx.

The first integral is solved with a u-substitution

u = x^2 - 6x + 13

, yielding

\ln|x^2 - 6x + 13|

. The quadratic is always positive, so the absolute value is not needed. The second integral is an arctangent form:

5 \cdot \frac{1}{2}\arctan\left(\frac{x-3}{2}\right)

. Combining these gives

\ln(x^2 - 6x + 13) + \frac{5}{2}\arctan\left(\frac{x-3}{2}\right) + C

Question 2

Evaluate the definite integral $\int_0^1 \frac{x^2}{x+1} dx$ .

$-\frac{1}{2}$
$\ln(2) + \frac{1}{2}$
$1 - \ln(2)$
$\ln(2) - \frac{1}{2}$ (correct answer)

Explanation: When you encounter a rational function where the degree of the numerator is close to the degree of the denominator, polynomial long division is often your best approach. Here, we have

\frac{x^2}{x+1}

, where both numerator and denominator are degree 1 apart. Dividing

x^2

(x+1)

$x^2 = (x+1) \cdot x - x$
$-x = (x+1) \cdot (-1) + 1$

\frac{x^2}{x+1} = x - 1 + \frac{1}{x+1}

Now we can integrate term by term:

\int_0^1 \frac{x^2}{x+1} dx = \int_0^1 \left(x - 1 + \frac{1}{x+1}\right) dx

= \left[\frac{x^2}{2} - x + \ln|x+1|\right]_0^1

= \left(\frac{1}{2} - 1 + \ln(2)\right) - \left(0 - 0 + \ln(1)\right)

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

Answer D is correct. Answer A gives

-\frac{1}{2}

, which would result from forgetting the logarithmic term entirely. Answer B gives

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

, a sign error when computing

\frac{1}{2} - 1

. Answer C gives

1 - \ln(2)

, likely from incorrectly handling the polynomial division or making multiple arithmetic errors. Study tip: When integrating rational functions, always check if polynomial long division simplifies the problem first. This technique transforms difficult integrals into manageable polynomial plus simple rational terms.

Question 3

To evaluate $\int \frac{dx}{ax^2+bx+c}$ where $b^2-4ac < 0$ and $a>0$ , completing the square transforms the denominator into the form $a[(x+k)^2+h^2]$ . What is $h$ in terms of $a, b, c$ ?

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

Explanation: When you encounter integrals with quadratic denominators where the discriminant is negative, completing the square is essential to transform the expression into a form that leads to an inverse tangent antiderivative. To complete the square for

ax^2 + bx + c

, you factor out the coefficient of

x^2

first:

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

. Then complete the square inside the parentheses by adding and subtracting

(\frac{b}{2a})^2

a\left[x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} - \frac{b^2}{4a^2} + \frac{c}{a}\right]

This becomes:

a\left[\left(x + \frac{b}{2a}\right)^2 + \frac{c}{a} - \frac{b^2}{4a^2}\right]

Simplifying the constant term:

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

So we have:

a\left[\left(x + \frac{b}{2a}\right)^2 + \frac{4ac - b^2}{4a^2}\right]

Comparing with

a[(x+k)^2 + h^2]

, we see that

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

, so

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

, which is choice B. Choice A uses

\sqrt{b^2 - 4ac}

, which would be imaginary since

b^2 - 4ac < 0

. Choice C gives you

k

, not

h

. Choice D has the wrong denominator—it should be

2a

, not

a

. Remember: when

b^2 - 4ac < 0

, always flip the sign under the square root to

4ac - b^2

during completing the square to ensure you're working with real numbers.

Question 4

For what value of $b > 0$ does $\int_0^b \frac{2x}{x^2+1} dx = \int_0^1 \frac{2x^3+2x+1}{x^2+1} dx$ ?

$e^{1+\pi/4} - 1$
$\sqrt{e^{\pi/4} - 1}$
$\sqrt{e - 1}$
$\sqrt{e^{1+\pi/4} - 1}$ (correct answer)

Explanation: When you encounter equations with definite integrals, you need to evaluate each integral separately and then solve for the unknown parameter. This question tests your ability to compute integrals and work with logarithmic equations. First, let's evaluate the left integral. For

\int_0^b \frac{2x}{x^2+1} dx

, notice that the derivative of

x^2+1

2x

, so this is a perfect setup for substitution. Let

u = x^2+1

, then

du = 2x\,dx

. This gives us

\int \frac{1}{u} du = \ln|u| = \ln(x^2+1)

. Evaluating from 0 to

b

\ln(b^2+1) - \ln(1) = \ln(b^2+1)

. For the right integral, split

\frac{2x^3+2x+1}{x^2+1}

by polynomial long division:

\frac{2x^3+2x+1}{x^2+1} = 2x + \frac{1}{x^2+1}

. So

\int_0^1 \left(2x + \frac{1}{x^2+1}\right) dx = [x^2 + \arctan(x)]_0^1 = 1 + \frac{\pi}{4}

. Setting the integrals equal:

\ln(b^2+1) = 1 + \frac{\pi}{4}

. Solving for

b

b^2+1 = e^{1+\pi/4}

, so

b^2 = e^{1+\pi/4} - 1

, giving

b = \sqrt{e^{1+\pi/4} - 1}

. Choice A omits the square root, choice B uses only

\pi/4

in the exponent (missing the 1), and choice C uses only

e

(missing the

\pi/4

). Study tip: When comparing definite integrals, evaluate each separately first, then solve the resulting equation. Watch for substitution opportunities when the numerator is the derivative of the denominator.

Question 5

Let $F(x) = \int \frac{x^4}{x^2+1} dx$ . If $F(0) = 1$ , what is the value of $F(1)$ ?

$\frac{1}{3} + \frac{\pi}{4}$ (correct answer)
$\frac{\pi}{4} - \frac{2}{3}$
$\frac{1}{3} + \frac{\pi}{4} + 1$
$-\frac{2}{3}$

Explanation: First, perform long division on

\frac{x^4}{x^2+1}

. A quick way is to rewrite the numerator:

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

. Now, integrate:

F(x) = \int (x^2-1 + \frac{1}{x^2+1}) dx = \frac{x^3}{3} - x + \arctan(x) + C

. Use the initial condition

F(0) = 1

\frac{0}{3} - 0 + \arctan(0) + C = 1 \implies C=1

. So,

F(x) = \frac{x^3}{3} - x + \arctan(x) + 1

. Finally, evaluate at

x=1

F(1) = \frac{1^3}{3} - 1 + \arctan(1) + 1 = \frac{1}{3} + \frac{\pi}{4}

Question 6

Which integral is generated by an appropriate u-substitution after preparing $\int \frac{dx}{\sqrt{3-4x-4x^2}}$ for integration?

$\frac{1}{2} \int \frac{1}{\sqrt{u^2+4}} du$
$\int \frac{1}{\sqrt{4-u^2}} du$
$\int \frac{1}{\sqrt{u^2-4}} du$
$\frac{1}{2} \int \frac{1}{\sqrt{4-u^2}} du$ (correct answer)

Explanation: When you encounter an integral with a quadratic expression under a square root, you need to complete the square first, then use u-substitution to transform it into a standard form. Starting with

\int \frac{dx}{\sqrt{3-4x-4x^2}}

, first factor out the coefficient of

x^2

from the quadratic:

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

. To complete the square for

x^2+x

, add and subtract

(\frac{1}{2})^2 = \frac{1}{4}

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

. Substituting back:

3-4x-4x^2 = -4[(x+\frac{1}{2})^2 - \frac{1}{4}] + 3 = -4(x+\frac{1}{2})^2 + 1 + 3 = 4 - 4(x+\frac{1}{2})^2

. The integral becomes

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

. Now substitute

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

, so

du = 2dx

dx = \frac{1}{2}du

. Also,

4(x+\frac{1}{2})^2 = u^2

. This gives us

\int \frac{\frac{1}{2}du}{\sqrt{4-u^2}} = \frac{1}{2}\int \frac{du}{\sqrt{4-u^2}}

, which matches answer D. Answer A has the wrong sign (

u^2+4

instead of

4-u^2

) and represents a different standard form. Answer B is missing the factor of

\frac{1}{2}

that comes from the u-substitution. Answer C has both the wrong sign under the square root and represents yet another standard form entirely. Study tip: Always complete the square first for quadratic expressions under radicals, then carefully track coefficients during u-substitution to avoid losing constant factors.

Question 7

Find the area of the region enclosed by the graph of $y = \frac{x^2+x+1}{x^2+1}$ , the x-axis, and the lines $x=0$ and $x=1$ .

$\frac{1}{2}\ln(2)$
$1 + \frac{\pi}{4}$
$1 + \frac{1}{2}\ln(2)$ (correct answer)
$1 + \ln(2)$

Explanation: When you encounter an area problem with a rational function, your first instinct should be to set up a definite integral. Here you need to find the area between the curve

y = \frac{x^2+x+1}{x^2+1}

and the x-axis from

x=0

x=1

. The key insight is recognizing that this rational function can be simplified using polynomial long division or by rewriting the numerator strategically. Notice that

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

, so:

\frac{x^2+x+1}{x^2+1} = \frac{x^2+1}{x^2+1} + \frac{x}{x^2+1} = 1 + \frac{x}{x^2+1}

Now the integral becomes much more manageable:

\int_0^1 \left(1 + \frac{x}{x^2+1}\right)dx = \int_0^1 1\,dx + \int_0^1 \frac{x}{x^2+1}dx

The first integral gives you

x\big|_0^1 = 1

. For the second, use the substitution

u = x^2+1

, so

du = 2x\,dx

, giving you

\frac{1}{2}\ln|x^2+1|\big|_0^1 = \frac{1}{2}[\ln(2) - \ln(1)] = \frac{1}{2}\ln(2)

. Therefore, the total area is

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

, which is answer C. Answer A gives only the logarithmic portion, missing the constant term. Answer B incorrectly involves

\frac{\pi}{4}

, which would appear if you mistakenly used

\arctan(x)

. Answer D doubles the logarithmic coefficient, likely from forgetting the factor of

\frac{1}{2}

in the substitution. Strategy tip: Always look for ways to decompose rational functions before integrating—polynomial long division or strategic rewriting can transform a difficult integral into simple, recognizable forms.

Question 8

The integral $\int \frac{x^n}{x-1} dx$ for an integer $n \geq 1$ results in a function of the form $P(x) + \ln|x-1| + C$ , where $P(x)$ is a polynomial. What is the degree of $P(x)$ ?

$n-1$
$n$ (correct answer)
$n+1$
$1$

Explanation: To integrate

\frac{x^n}{x-1}

, we perform polynomial long division. The division of

x^n

x-1

results in a quotient and a remainder. Using the remainder theorem, the remainder is

(1)^n=1

. The quotient,

Q(x)

, will have a degree of

n-1

\frac{x^n}{x-1} = Q(x) + \frac{1}{x-1}

where

Q(x) = x^{n-1} + x^{n-2} + \dots + x + 1

.
The integral is then:

\int \left(x^{n-1} + x^{n-2} + \dots + x + 1 + \frac{1}{x-1}\right) dx

The term

P(x)

in the result corresponds to the integral of the polynomial

Q(x)

P(x) = \int Q(x) dx = \int (x^{n-1} + x^{n-2} + \dots + 1) dx

P(x) = \frac{x^n}{n} + \frac{x^{n-1}}{n-1} + \dots + x

The highest power of

x

P(x)

n

, so the degree of the polynomial

P(x)

n

Question 9

Consider the integral $I = \int \frac{P(x)}{Q(x)} dx$ , where $P(x)$ is a polynomial of degree 4, $Q(x)$ is a polynomial of degree 2, and $Q(x)$ is irreducible. Which sequence of steps is the most direct and appropriate method to evaluate $I$ ?

Perform long division, then complete the square on the denominator of the resulting rational term and split this term into logarithmic and arctangent parts for integration. (correct answer)
Apply partial fraction decomposition directly to the integrand $\frac{P(x)}{Q(x)}$ , expressing it as a sum of simpler rational functions.
Complete the square in the denominator $Q(x)$ first, then attempt to simplify the entire rational expression before integrating.
Use a single u-substitution with $u = Q(x)$ and transform the entire integral in terms of $u$ before proceeding.

Explanation: Since the degree of the numerator

P(x)

(4) is greater than or equal to the degree of the denominator

Q(x)

(2), the first required step is polynomial long division. This will result in an expression of the form

\text{Quotient} + \frac{\text{Remainder}}{Q(x)}

, where the degree of the Remainder is less than 2. The Quotient is a polynomial and can be integrated easily. The term

\frac{\text{Remainder}}{Q(x)}

has an irreducible quadratic denominator, so the next step is to complete the square in

Q(x)

. After completing the square, the fraction is typically split into two parts: one that integrates to a natural logarithm and one that integrates to an arctangent.

Question 10

Let $f(x) = \frac{1}{x^2-4x+5}$ and $g(x) = \frac{1}{\sqrt{2x-x^2}}$ . Let $A = \int_2^3 f(x)dx$ and $B = \int_1^{1.5} g(x)dx$ . Which of the following describes the relationship between A and B?

$A = \frac{\pi}{6}, B = \frac{\pi}{4}$
$A = \frac{\pi}{4}, B = \frac{\pi}{6}$ (correct answer)
$A = \arctan(3), B = \arcsin(0.5)$
$A = \ln(2), B = \ln(1.5)$

Explanation: When you encounter integrals with expressions like

\frac{1}{x^2-4x+5}

\frac{1}{\sqrt{2x-x^2}}

, look for opportunities to complete the square and use inverse trigonometric function formulas. For integral A, complete the square in the denominator:

x^2-4x+5 = (x-2)^2+1

. So

A = \int_2^3 \frac{1}{(x-2)^2+1}dx

. Using substitution

u = x-2

, when

x=2

u=0

; when

x=3

u=1

. This gives us

A = \int_0^1 \frac{1}{u^2+1}du = \arctan(u)\Big|_0^1 = \arctan(1) - \arctan(0) = \frac{\pi}{4}

. For integral B, complete the square under the square root:

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

. So

B = \int_1^{1.5} \frac{1}{\sqrt{1-(x-1)^2}}dx

. Using substitution

u = x-1

, when

x=1

u=0

; when

x=1.5

u=0.5

. This gives us

B = \int_0^{0.5} \frac{1}{\sqrt{1-u^2}}du = \arcsin(u)\Big|_0^{0.5} = \arcsin(0.5) - \arcsin(0) = \frac{\pi}{6}

. Therefore

A = \frac{\pi}{4}

and

B = \frac{\pi}{6}

, making B correct. Choice A reverses these values. Choice C stops at the intermediate step before evaluating the special angles. Choice D incorrectly suggests these are logarithmic integrals rather than inverse trigonometric ones. Remember: expressions of the form

\frac{1}{u^2+1}

integrate to

\arctan(u)

, while

\frac{1}{\sqrt{1-u^2}}

integrates to

\arcsin(u)

. Complete the square first to recognize these patterns.

Question 11

A certain calculation requires finding the antiderivative of $f(x) = \frac{x+1}{4x^2+1}$ . Another calculation requires finding the antiderivative of $g(x) = \frac{x^3}{4x^2+1}$ . Which of the following statements is true?

Evaluating $\int f(x) dx$ requires long division, while $\int g(x) dx$ does not.
Evaluating $\int g(x) dx$ requires long division, while $\int f(x) dx$ does not. (correct answer)
Both integrals require long division as a first step.
Neither integral requires long division, they can both be solved by direct substitution.

Explanation: When you encounter rational functions where the degree of the numerator is greater than or equal to the degree of the denominator, you need to consider whether polynomial long division is necessary before integration. For

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

, the numerator has degree 1 and denominator has degree 2, so the degree of the numerator is less than the denominator. This means you can integrate directly using substitution and partial fraction techniques. You can split this as

\int \frac{x}{4x^2+1} dx + \int \frac{1}{4x^2+1} dx

. The first integral uses u-substitution with

u = 4x^2+1

, and the second is an arctangent form. For

\int g(x) dx = \int \frac{x^3}{4x^2+1} dx

, the numerator has degree 3 and denominator has degree 2. Since the numerator's degree is higher, you must perform polynomial long division first. Dividing

x^3

4x^2+1

gives you

\frac{1}{4}x - \frac{1}{16} + \frac{\frac{1}{16}}{4x^2+1}

, which you can then integrate term by term. Option A incorrectly states that

f(x)

requires long division when it doesn't. Option C is wrong because only

g(x)

needs long division. Option D is incorrect because

g(x)

cannot be solved by direct substitution alone due to the degree issue. Study tip: Always compare the degrees of numerator and denominator first. If the numerator's degree is greater than or equal to the denominator's degree, perform long division before attempting integration.

Question 12

The expression $\frac{x^4+3x^2+1}{x^2+1}$ can be simplified to $P(x) + \frac{C}{x^2+1}$ where $P(x)$ is a polynomial and $C$ is a constant. Find $\int_0^{\sqrt{3}} \frac{x^4+3x^2+1}{x^2+1} dx$ .

$3\sqrt{3} + \frac{\pi}{3}$
$3\sqrt{3} - \frac{\pi}{3}$ (correct answer)
$\sqrt{3} - \frac{\pi}{3}$
$3\sqrt{3} - \frac{\pi}{6}$

Explanation: When you encounter a rational function where the degree of the numerator is greater than or equal to the degree of the denominator, you need to perform polynomial long division first. This separates the function into a polynomial part plus a simpler rational remainder. Let's divide

x^4 + 3x^2 + 1

x^2 + 1

. Since there's no

x^3

term in the numerator, we get:

$x^4 ÷ x^2 = x^2$ , giving $x^2(x^2 + 1) = x^4 + x^2$
Subtracting: $(x^4 + 3x^2 + 1) - (x^4 + x^2) = 2x^2 + 1$
$2x^2 ÷ x^2 = 2$ , giving $2(x^2 + 1) = 2x^2 + 2$
Subtracting: $(2x^2 + 1) - (2x^2 + 2) = -1$

\frac{x^4+3x^2+1}{x^2+1} = x^2 + 2 + \frac{-1}{x^2+1}

Now integrate:

\int_0^{\sqrt{3}} \left(x^2 + 2 - \frac{1}{x^2+1}\right) dx

This becomes:

\left[\frac{x^3}{3} + 2x - \arctan(x)\right]_0^{\sqrt{3}}

Evaluating:

\left(\frac{(\sqrt{3})^3}{3} + 2\sqrt{3} - \arctan(\sqrt{3})\right) - (0 + 0 - 0)

= \frac{3\sqrt{3}}{3} + 2\sqrt{3} - \frac{\pi}{3} = \sqrt{3} + 2\sqrt{3} - \frac{\pi}{3} = 3\sqrt{3} - \frac{\pi}{3}

Answer B is correct. Answer A adds instead of subtracts the arctangent term. Answer C miscalculates the polynomial integration. Answer D uses

\frac{\pi}{6}

instead of

\frac{\pi}{3}

for

\arctan(\sqrt{3})

. Study tip: Always perform polynomial long division when the numerator's degree isn't smaller than the denominator's, and remember that

\arctan(\sqrt{3}) = \frac{\pi}{3}

Question 13

Which expression represents the integral $\int \frac{1}{2x^2 + 4x + 10} dx$ ?

$\frac{1}{4}\arctan\left(\frac{x+1}{2}\right) + C$ (correct answer)
$\frac{1}{2}\arctan\left(\frac{x+1}{2}\right) + C$
$\frac{1}{2}\arctan(x+1) + C$
$\frac{1}{4}\ln(2x^2+4x+10) + C$

Explanation: First, factor out the leading coefficient from the denominator:

\int \frac{1}{2(x^2 + 2x + 5)} dx = \frac{1}{2} \int \frac{1}{x^2 + 2x + 5} dx

. Next, complete the square for the quadratic in the denominator:

x^2 + 2x + 5 = (x^2 + 2x + 1) + 4 = (x+1)^2 + 2^2

. The integral becomes

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

. This is an arctangent form

\frac{1}{a}\arctan(\frac{u}{a})

with

u=x+1

and

a=2

. Integrating gives:

\frac{1}{2} \left[ \frac{1}{2} \arctan\left(\frac{x+1}{2}\right) \right] + C = \frac{1}{4} \arctan\left(\frac{x+1}{2}\right) + C

Question 14

Find the area of the region bounded by the graph of $f(x) = \frac{1}{x^2 - 4x + 13}$ , the x-axis, and the lines $x=2$ and $x=5$ .

$\frac{\pi}{4}$
$\ln(2)$
$\frac{\pi}{9}$
$\frac{\pi}{12}$ (correct answer)

Explanation: The area is given by the definite integral

A = \int_{2}^{5} \frac{1}{x^2 - 4x + 13} dx

. The denominator is an irreducible quadratic, so we complete the square:

x^2 - 4x + 13 = (x^2 - 4x + 4) + 9 = (x-2)^2 + 3^2

The integral becomes

\int_{2}^{5} \frac{1}{(x-2)^2 + 3^2} dx

. Let

u = x-2

, so

du = dx

. The limits of integration change from

x=2, 5

u=0, 3

A = \int_{0}^{3} \frac{1}{u^2 + 3^2} du = \left[ \frac{1}{3}\arctan\left(\frac{u}{3}\right) \right]_{0}^{3}

Evaluating at the limits gives:

\frac{1}{3}\arctan\left(\frac{3}{3}\right) - \frac{1}{3}\arctan\left(\frac{0}{3}\right) = \frac{1}{3}\arctan(1) - 0 = \frac{1}{3} \cdot \frac{\pi}{4} = \frac{\pi}{12}

Question 15

Evaluate the integral $\int_{-3}^{1} \frac{1}{x^2+2x+5} dx$ .

$\frac{\pi}{2}$
$0$
$\frac{\pi}{8}$
$\frac{\pi}{4}$ (correct answer)

Explanation: The denominator

x^2+2x+5

has no real roots, so we complete the square.

x^2+2x+5 = (x^2+2x+1)+4 = (x+1)^2+2^2

The integral is

\int_{-3}^{1} \frac{1}{(x+1)^2+2^2} dx

. We use the substitution

u=x+1

, so

du=dx

. The limits change from

x=-3, 1

u=-2, 2

\int_{-2}^{2} \frac{1}{u^2+2^2} du = \left[ \frac{1}{2}\arctan\left(\frac{u}{2}\right) \right]_{-2}^{2}

Evaluating at the limits:

\frac{1}{2}\arctan\left(\frac{2}{2}\right) - \frac{1}{2}\arctan\left(\frac{-2}{2}\right) = \frac{1}{2}\arctan(1) - \frac{1}{2}\arctan(-1)

= \frac{1}{2}\left(\frac{\pi}{4}\right) - \frac{1}{2}\left(-\frac{\pi}{4}\right) = \frac{\pi}{8} + \frac{\pi}{8} = \frac{\pi}{4}

Alternatively, recognizing that

\frac{1}{u^2+4}

is an even function, the integral is

2 \int_{0}^{2} \frac{1}{u^2+4} du = 2 \left[ \frac{1}{2}\arctan(\frac{u}{2}) \right]_{0}^{2} = \arctan(1) - \arctan(0) = \frac{\pi}{4}

Question 16

Which of the following is equivalent to the indefinite integral $\int \frac{x^3 + 2x^2 + 6x + 5}{x^2 + 2x + 2} dx$ ?

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

Explanation: The degree of the numerator is greater than the degree of the denominator, so we use long division first.

\frac{x^3 + 2x^2 + 6x + 5}{x^2 + 2x + 2} = x + \frac{4x+5}{x^2+2x+2}

The integral becomes

\int x\,dx + \int \frac{4x+5}{x^2+2x+2} dx

. The first term is

\frac{x^2}{2}

. For the second term, we split the fraction. The derivative of the denominator is

2x+2

. We create this in the numerator:

4x+5 = 2(2x+2) + 1

So,

\int \frac{4x+5}{x^2+2x+2} dx = \int \frac{2(2x+2)}{x^2+2x+2} dx + \int \frac{1}{x^2+2x+2} dx

. The first part is

2\ln(x^2+2x+2)

. For the second part, complete the square:

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

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

Combining all parts gives

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

Question 17

Which of the following is equivalent to $\int \frac{2x-3}{x^2+6x+10} dx$ ?

$\ln(x^2+6x+10) + 9\arctan(x+3) + C$
$\ln(x^2+6x+10) - 3\arctan(x+3) + C$
$\ln(x^2+6x+10) - 9\arctan(x+3) + C$ (correct answer)
$\frac{1}{2}\ln(x^2+6x+10) - 9\arctan(x+3) + C$

Explanation: The integrand's denominator is an irreducible quadratic. The derivative of

x^2+6x+10

2x+6

. We manipulate the numerator to contain this expression:

2x-3 = (2x+6) - 9

This allows us to split the integral into two parts:

\int \frac{2x+6}{x^2+6x+10} dx - \int \frac{9}{x^2+6x+10} dx

The first integral is a standard natural log form, yielding

\ln(x^2+6x+10)

(absolute value is not needed as the quadratic is always positive). For the second integral, we complete the square in the denominator:

x^2+6x+10 = (x^2+6x+9)+1 = (x+3)^2+1

So the second integral is

-9 \int \frac{1}{(x+3)^2+1} dx = -9\arctan(x+3)

.
Combining the parts gives

\ln(x^2+6x+10) - 9\arctan(x+3) + C

Question 18

Let $f(x) = \frac{2x^3 - 4x^2 + 3x - 5}{x-2}$ . The integral $\int f(x) dx$ can be written as $G(x) + C$ . Which of the following is $G(x)$ ?

$\frac{2}{3}x^3 - 4x^2 + 19x - 43\ln|x+2|$
$\frac{2}{3}x^3 + 3x$
$\frac{2}{3}x^3 + 3x - (x-2)^{-1}$
$\frac{2}{3}x^3 + 3x + \ln|x-2|$ (correct answer)

Explanation: To integrate

f(x)

, we first simplify it using polynomial division. Since the denominator is of the form

x-c

, we can use synthetic division with

c=2

. The coefficients of the numerator are

2, -4, 3, -5

. \

\begin{array}{c|cccc} 2 & 2 & -4 & 3 & -5 \\ & & 4 & 0 & 6 \\ \hline & 2 & 0 & 3 & 1 \end{array} $$ \ This means that $\frac{2x^3 - 4x^2 + 3x - 5}{x-2} = 2x^2 + 0x + 3 + \frac{1}{x-2}$. \ Now, we integrate this expression: \ $$ \int \left( 2x^2 + 3 + \frac{1}{x-2} \right) dx = \frac{2x^3}{3} + 3x + \ln|x-2| $$ \ Thus, $G(x) = \frac{2}{3}x^3 + 3x + \ln|x-2|$.

Question 19

For what values of the constant $b$ does the integral $\int \frac{1}{x^2+bx+1} dx$ result in an expression involving an arctangent function, rather than a natural logarithm?

$-2 < b < 2$ (correct answer)
$b = 2$ or $b = -2$
All real values of $b$
$b > 2$ or $b < -2$

Explanation: When you encounter an integral of the form

\int \frac{1}{x^2+bx+1} dx

, the key is recognizing that the result depends on whether the quadratic denominator can be factored into real linear terms or not. This determines whether you'll use partial fractions (leading to logarithms) or complete the square (leading to arctangent). The quadratic

x^2+bx+1

factors into real linear terms when its discriminant is positive:

b^2-4(1)(1) > 0

, or

b^2 > 4

. When this happens, you can use partial fractions, resulting in natural logarithms. However, when the discriminant is negative (

b^2 < 4

), the quadratic has no real roots, so you complete the square and get an arctangent form. For an arctangent result, you need

b^2 < 4

, which means

-2 < b < 2

. This makes choice A correct. Choice B (

b = ±2

) represents the boundary cases where

b^2 = 4

, giving a perfect square denominator and resulting in a simple rational function, not arctangent. Choice C incorrectly suggests the form doesn't depend on

b

's value. Choice D (

b > 2

b < -2

) gives the conditions for logarithmic results, which is the opposite of what we want. Study tip: Remember the discriminant rule for rational function integration: negative discriminant → arctangent, positive discriminant → logarithm. The boundary between these regions (where the discriminant equals zero) often appears as a distractor choice.

Question 20

A particle's velocity is given by $v(t) = \frac{1}{\sqrt{8+2t-t^2}}$ for $t$ in the interval $(-2, 4)$ . Find the displacement of the particle from $t=0$ to $t=2$ .

$2\arcsin(\frac{1}{3})$ (correct answer)
$\arcsin(\frac{1}{3})$
$0$
$\frac{2}{3}\arcsin(\frac{1}{3})$

Explanation: Displacement is the integral of velocity, so we must evaluate

\int_{0}^{2} \frac{1}{\sqrt{8+2t-t^2}} dt

. To handle the quadratic in the square root, we complete the square:

8+2t-t^2 = -(t^2 - 2t - 8) = -( (t^2 - 2t + 1) - 9 ) = -((t-1)^2 - 9) = 9-(t-1)^2

The integral becomes

\int_{0}^{2} \frac{1}{\sqrt{9-(t-1)^2}} dt

. We use the substitution

u = t-1

, so

du = dt

. The limits of integration become

u(0)= -1

and

u(2) = 1

\int_{-1}^{1} \frac{1}{\sqrt{9-u^2}} du = \int_{-1}^{1} \frac{1}{\sqrt{3^2-u^2}} du

This is the standard form for arcsin:

\left[ \arcsin\left(\frac{u}{3}\right) \right]_{-1}^{1} = \arcsin\left(\frac{1}{3}\right) - \arcsin\left(-\frac{1}{3}\right)

Since

\arcsin(-x) = -\arcsin(x)

, this is

\arcsin(\frac{1}{3}) - (-\arcsin(\frac{1}{3})) = 2\arcsin(\frac{1}{3})