Calculus 2 Quiz: Partial Fractions Repeated And Quadratics

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Partial Fractions Repeated And QuadraticsQuestion 1 of 20

Given the identity $\frac{1}{x(x-1)^2} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{(x-1)^2}$ , find the value of B.

-1

-2

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Calculus 2 Quiz

Calculus 2 Quiz: Partial Fractions Repeated And Quadratics

Practice Partial Fractions Repeated And Quadratics 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 Partial Fractions Repeated And Quadratics, 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

Given the identity $\frac{1}{x(x-1)^2} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{(x-1)^2}$ , find the value of B.

1
-1 (correct answer)
0
-2

Explanation: Start with

1 = A(x-1)^2 + Bx(x-1) + Cx

. Use the cover-up method for the non-repeated and highest power repeated factors. Let

x=0

1 = A(-1)^2 \Rightarrow A=1

. Let

x=1

1 = C(1) \Rightarrow C=1

. To find B, we can equate the coefficients of

x^2

. On the right side, the

x^2

terms are

Ax^2

and

Bx^2

. So,

A+B

must equal the

x^2

coefficient on the left, which is 0.

A+B=0 \Rightarrow 1+B=0 \Rightarrow B=-1

Question 2

Evaluate the integral $\int \frac{2x}{(x+1)^2} dx$ .

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

Explanation: Using partial fractions, we set

\frac{2x}{(x+1)^2} = \frac{A}{x+1} + \frac{B}{(x+1)^2}

. Multiplying by the denominator gives

2x = A(x+1) + B

. Setting

x=-1

gives

B=-2

. Setting

x=0

gives

0 = A+B \Rightarrow A=-B=2

. The integral becomes

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

. This integrates to

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

. A common error is a sign mistake in the integration of

(x+1)^{-2}

Question 3

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

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

Explanation: The denominator

x^2+2x+5

is an irreducible quadratic. Complete the square:

(x+1)^2 + 4

. The derivative of the denominator is

2x+2

. We rewrite the numerator to contain this term:

2x+3 = (2x+2) + 1

. The integral becomes

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

. The first integral yields

\ln(x^2+2x+5)

(absolute value is not needed as the quadratic is always positive). For the second integral, use substitution

u=x+1

a=2

, yielding

\frac{1}{2}\arctan\left(\frac{x+1}{2}\right)

. Combining these gives the correct answer.

Question 4

The integration of a rational function $f(x) = \frac{P(x)}{Q(x)}$ using partial fractions results in the expression $\frac{-2}{x+3} + \ln(x^2+9) + C$ . Which of the following could be the denominator $Q(x)$ of the original function $f(x)$ ?

$(x+3)(x^2+9)$
$(x+3)^2(x^2+9)$ (correct answer)
$(x+3)^2(x+9)$
$(x-3)(x+3)(x^2+9)$

Explanation: We analyze the terms in the antiderivative to deduce the form of the partial fraction decomposition. The term

\ln(x^2+9)

results from integrating a term like

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

. The term

\frac{-2}{x+3}

results from integrating

\frac{A}{(x+3)^2}

(specifically, where A=-2). This implies the decomposition must have contained terms with denominators

(x+3)^2

and

(x^2+9)

. The original denominator

Q(x)

must be the product of these factors, which is

(x+3)^2(x^2+9)

Question 5

The evaluation of $\int \frac{P(x)}{Q(x)} dx$ via partial fractions yields an antiderivative containing the term $\frac{1}{2} \arctan\left(\frac{x-1}{2}\right)$ . Which factor must be part of the denominator $Q(x)$ ?

$x^2+4$
$x^2 - 2x + 5$ (correct answer)
$x^2 - 2x + 1$
$4x^2 - 8x + 20$

Explanation: The integral form

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

arises from integrating

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

. In this case,

a=2

and the argument of arctan suggests a substitution

u=x-1

. This means the original integral must have contained a term of the form

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

for some constant k. Expanding the denominator gives

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

. Therefore,

x^2-2x+5

must be a factor of

Q(x)

Question 6

A student computes $\int \frac{x^3}{(x^2+4)^2} dx$ by first setting up the partial fraction decomposition $\frac{Ax+B}{x^2+4} + \frac{Cx+D}{(x^2+4)^2}$ . While their final answer may be correct, what is the primary conceptual error or inefficiency in this approach?

The student used partial fractions when a direct u-substitution would have been simpler. (correct answer)
The decomposition is incorrect; it should only have one term $\frac{Cx+D}{(x^2+4)^2}$ .
The integral of the $\frac{Cx+D}{(x^2+4)^2}$ term requires integration by parts, not u-substitution.
The factor $x^2+4$ is reducible, so the decomposition form is fundamentally wrong.

Explanation: While partial fractions will work, it is not the most efficient method. This integral is solved more directly with the substitution

u = x^2+4

. Then

x^2 = u-4

and

du = 2x dx

, so

x dx = du/2

. The integral becomes

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

, which is easily integrated. Recognizing this simpler path is a key skill. The use of a more complex method when a simpler one exists represents a conceptual inefficiency.

Question 7

Consider the integral $\int \frac{3x^2+2}{x(x^2+1)^2} dx$ . After setting up the partial fraction decomposition as $\frac{A}{x} + \frac{Bx+C}{x^2+1} + \frac{Dx+E}{(x^2+1)^2}$ , which of the following integration techniques is essential for finding the antiderivative of the term $\frac{E}{(x^2+1)^2}$ ?

A simple u-substitution with $u=x^2+1$ .
Integration by parts with $u = E$ .
The Heaviside cover-up method.
A trigonometric substitution such as $x = \tan(\theta)$ . (correct answer)

Explanation: The term

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

cannot be solved with a simple u-substitution. This integral is a classic case for trigonometric substitution. Let

x = \tan(\theta)

, so

dx = \sec^2(\theta) d\theta

and

x^2+1 = \tan^2(\theta)+1 = \sec^2(\theta)

. The integral becomes

\int \frac{E}{(\sec^2(\theta))^2} \sec^2(\theta) d\theta = E \int \frac{1}{\sec^2(\theta)} d\theta = E \int \cos^2(\theta) d\theta

, which can then be solved using a power-reducing identity.

Question 8

Integrating $\int \frac{1}{x(x^2-2x+5)} dx$ using partial fractions leads to an antiderivative with three distinct functional forms. Two of them are a natural logarithm and an inverse tangent. What is the nature of the third functional form?

Another inverse tangent with a different argument.
A rational function of the form $c/x$ .
A power function of the form $cx^n$ where $n \neq -1$ .
Another natural logarithm with a different argument. (correct answer)

Explanation: When you encounter a rational function with polynomial factors in the denominator, partial fraction decomposition breaks it into simpler fractions that you can integrate term by term. For

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

, you'll decompose this as

\frac{A}{x} + \frac{Bx+C}{x^2-2x+5}

. The first term

\frac{A}{x}

integrates to give you a natural logarithm:

A \ln|x|

. The quadratic

x^2-2x+5

has discriminant

(-2)^2 - 4(1)(5) = -16 < 0

, so it's irreducible with no real roots. When you integrate

\frac{Bx+C}{x^2-2x+5}

, you'll need to split this into two parts. One part (from the

Bx

term) will involve substitution and yield another natural logarithm of the form

\ln|x^2-2x+5|

. The other part (from the constant

C

) requires completing the square:

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

, leading to an inverse tangent function. Looking at the wrong answers: A) suggests another inverse tangent, but the partial fraction setup only produces one arctangent term from the irreducible quadratic. B) claims a rational function

c/x

, but this would be a derivative, not an antiderivative from our integral. C) proposes a power function

cx^n

, but none of our partial fractions integrate to produce such terms. The correct answer is D) because you get two distinct logarithmic terms: one from

\frac{A}{x}

and another from integrating the linear numerator part of the irreducible quadratic fraction. Study tip: When decomposing rational functions, irreducible quadratics always produce both logarithmic and inverse tangent terms in the final antiderivative.

Question 9

What is $\int \frac{x}{(x^2 + 1)^2} dx$ ?

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

Explanation: When you encounter an integral with the form

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

, this is a perfect candidate for u-substitution because the numerator

x

is related to the derivative of the expression in the denominator. Let

u = x^2 + 1

. Then

du = 2x \, dx

, which means

x \, dx = \frac{1}{2} du

. Substituting this into our integral:

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

Using the power rule for integration:

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

Substituting back

u = x^2 + 1

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

This confirms answer C is correct. Answer A,

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

, results from incorrectly keeping

x^2

in the numerator instead of recognizing the u-substitution pattern. Answer B,

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

, has the right form but the wrong sign—this comes from forgetting that

u^{-2}

integrates to

u^{-1}/(-1)

, not

u^{-1}/(1)

. Answer D,

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

, combines both the sign error and the incorrect numerator. Study tip: When you see

x

in the numerator and

(x^2 + \text{constant})

in the denominator, immediately think u-substitution with

u = x^2 + \text{constant}

. The

x

will become part of your

du

Question 10

In evaluating $\int \frac{2x^2 - x + 3}{(x - 1)^2(x^2 + x + 1)} dx$ , after finding that the partial fraction decomposition is $\frac{2}{x - 1} + \frac{1}{(x - 1)^2} + \frac{-2x + 1}{x^2 + x + 1}$ , what is the next essential step for integrating the term $\frac{-2x + 1}{x^2 + x + 1}$ ?

Complete the square in the denominator to get $(x + \frac{1}{2})^2 + \frac{3}{4}$
Factor the denominator as $(x + \frac{1}{2})^2 - \frac{1}{4} + 1$
Use integration by parts with $u = -2x + 1$
Split into $\frac{-2x}{x^2 + x + 1} + \frac{1}{x^2 + x + 1}$ and handle each part separately (correct answer)

Explanation: When you encounter a rational function with an irreducible quadratic in the denominator after partial fraction decomposition, the key strategy is to split the numerator to create terms that match standard integral forms. For

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

, you should split this into

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

as shown in option D. This separation is strategic: the first term

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

will integrate using substitution (since the numerator is related to the derivative of the denominator), while the second term

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

will require completing the square and result in an arctangent form. Option A describes completing the square, which is indeed necessary, but only for the second term after splitting. It's not the immediate next step for the entire fraction. Option B shows an incorrect completion of the square. When you complete the square for

x^2 + x + 1

, you get

(x + \frac{1}{2})^2 + \frac{3}{4}

, not the form shown. Option C suggests integration by parts, which is inappropriate here. Integration by parts is used for products of functions, not rational functions with quadratic denominators. The splitting strategy in D allows you to handle each piece with the appropriate technique: substitution for the first term and trigonometric substitution (after completing the square) for the second. Remember: when dealing with linear numerators over irreducible quadratics, always split first to separate the "substitution part" from the "arctangent part."

Question 11

In the partial fraction decomposition $\frac{3x^2 - x + 2}{(x - 1)^2(x^2 + x + 1)} = \frac{A}{x - 1} + \frac{B}{(x - 1)^2} + \frac{Cx + D}{x^2 + x + 1}$ , which method would be most efficient for finding coefficient $A$ ?

Multiply both sides by $(x - 1)$ , then substitute $x = 1$
Multiply both sides by $(x - 1)^2$ , take the derivative, then substitute $x = 1$ (correct answer)
Clear denominators and compare coefficients of like terms
Substitute $x = 0$ after clearing denominators

Explanation: For coefficient

A

in a repeated linear factor, we use the derivative method: multiply by

(x-1)^2

to get

\frac{3x^2 - x + 2}{x^2 + x + 1} = A(x-1) + B + \frac{(Cx + D)(x-1)^2}{x^2 + x + 1}

. Taking the derivative eliminates the

B

term and makes the last term vanish at

x = 1

, leaving only the

A

term. Choice A would give a combination of

A

and

B

. Choice C works but is much more tedious. Choice D gives a combination of all coefficients and doesn't isolate

A

Question 12

Consider the partial fraction decomposition $\frac{P(x)}{(x^2 + 4)^2(x + 2)^3} = \frac{Ax + B}{x^2 + 4} + \frac{Cx + D}{(x^2 + 4)^2} + \frac{E}{x + 2} + \frac{F}{(x + 2)^2} + \frac{G}{(x + 2)^3}$ . If all coefficients $A, B, C, D, E, F, G$ are nonzero, what is the minimum possible degree of polynomial $P(x)$ ?

Degree 4
Degree 5
Degree 6 (correct answer)
Degree 7

Explanation: When you encounter partial fraction decomposition problems asking about polynomial degrees, you need to understand the relationship between the decomposition form and the original rational function's numerator. The given decomposition has denominators

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

, so this must equal the denominator of the original fraction. To find the minimum degree of

P(x)

, you need to determine when all coefficients can be nonzero simultaneously. The key insight is that

P(x)

must have degree less than the denominator's degree for proper partial fractions. The denominator

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

has degree

2 \cdot 2 + 1 \cdot 3 = 7

. So

P(x)

has maximum degree 6. For all coefficients to be nonzero,

P(x)

needs enough "complexity" to generate each term when decomposed. If

P(x)

had too low a degree, some coefficients would necessarily be zero. The minimum occurs when

P(x)

has degree 6, making this a proper fraction where the partial fraction decomposition can have all nonzero coefficients. Choice A (degree 4) is too restrictive - a degree 4 polynomial typically cannot generate all seven nonzero coefficients in this decomposition. Choice B (degree 5) similarly lacks sufficient terms. Choice D (degree 7) would make this an improper fraction, requiring polynomial long division first, which changes the problem structure entirely. Remember: for partial fractions with multiple repeated factors, the numerator polynomial needs sufficient degree (close to the denominator's degree) to ensure all coefficients can be nonzero. Always check that your fraction is proper before decomposing.

Question 13

To integrate $\int \frac{3x^3 + 2x^2 - x + 4}{(x^2 + 1)^2(x - 2)} dx$ using partial fractions, which decomposition form is correct?

$\frac{Ax + B}{x^2 + 1} + \frac{Cx + D}{(x^2 + 1)^2} + \frac{E}{x - 2}$ (correct answer)
$\frac{A}{x^2 + 1} + \frac{B}{(x^2 + 1)^2} + \frac{C}{x - 2}$
$\frac{Ax + B}{x^2 + 1} + \frac{C}{(x^2 + 1)^2} + \frac{D}{x - 2}$
$\frac{A}{x^2 + 1} + \frac{Bx + C}{(x^2 + 1)^2} + \frac{D}{x - 2}$

Explanation: For an irreducible quadratic factor

x^2 + 1

that appears squared, we need

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

for the first power and

\frac{Cx + D}{(x^2 + 1)^2}

for the second power. The linear factor

(x - 2)

gives

\frac{E}{x - 2}

. Choice B uses constants instead of linear terms for irreducible quadratics. Choice C omits the linear term in the repeated quadratic. Choice D incorrectly places the linear term only in the repeated factor.

Question 14

To evaluate $\int \frac{x^4+x^2-1}{x^3+x} dx$ , what is the first essential step and the resulting expression to be integrated?

Perform polynomial long division, resulting in $\int (x - \frac{1}{x(x^2+1)}) dx$ . (correct answer)
Factor the denominator and set up the decomposition $\frac{A}{x} + \frac{Bx+C}{x^2+1}$ .
Perform polynomial long division, resulting in $\int (x + \frac{2x^2-1}{x^3+x}) dx$ .
Use substitution $u=x^3+x$ , which transforms the integral into a simpler form.

Explanation: The degree of the numerator (4) is greater than the degree of the denominator (3), so the first step must be polynomial long division. Dividing

x^4+x^2-1

x^3+x

gives a quotient of

x

and a remainder of

-1

. Therefore, the integrand can be rewritten as

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

. Factoring the denominator of the remainder gives

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

. The resulting integral is

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

Question 15

A student sets up the partial fraction decomposition for $\frac{1}{(x^2+4)(x^2-4)}$ as $\frac{Ax+B}{x^2+4} + \frac{Cx+D}{x^2-4}$ . What is the fundamental error in this setup?

The numerator for the $x^2+4$ term should be a constant $A$ , not $Ax+B$ .
Both quadratic factors are reducible and should be factored further.
The term $x^2-4$ is a reducible quadratic and should be factored into $(x-2)(x+2)$ . (correct answer)
Long division should have been performed before attempting to decompose the fraction.

Explanation: The procedure for partial fractions requires the denominator to be factored as much as possible over the real numbers. The factor

x^2+4

is an irreducible quadratic. However, the factor

x^2-4

is a difference of squares and is reducible to

(x-2)(x+2)

. Therefore, the decomposition should be based on the factors

(x^2+4)

(x-2)

, and

(x+2)

, leading to the form

\frac{Ax+B}{x^2+4} + \frac{C}{x-2} + \frac{D}{x+2}

Question 16

For the partial fraction decomposition of $\frac{x^2}{(x^2+1)^2} = \frac{Ax+B}{x^2+1} + \frac{Cx+D}{(x^2+1)^2}$ , what is the value of D?

1
0
-1 (correct answer)
-2

Explanation: Begin with the identity

x^2 = (Ax+B)(x^2+1) + (Cx+D)

. Expand the right side:

x^2 = Ax^3 + Ax + Bx^2 + B + Cx + D

. Group by powers of x:

0x^3 + 1x^2 + 0x + 0 = Ax^3 + Bx^2 + (A+C)x + (B+D)

. Equating the coefficients for each power of x gives a system of equations:

A=0

B=1

A+C=0

, and

B+D=0

. From

A+C=0

, we get

C=0

. From

B+D=0

, we get

1+D=0

, which means

D=-1

Question 17

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

0 (correct answer)
$\ln(3/2) - 1/3$
$\ln(3) + 2/3$
$2\ln(3) - 1$

Explanation: This problem can be solved by recognizing the properties of the integrand. Let

f(x) = \frac{x^3}{(x^2+2)^2}

. We check if the function is even or odd.

f(-x) = \frac{(-x)^3}{((-x)^2+2)^2} = \frac{-x^3}{(x^2+2)^2} = -f(x)

. Since

f(x)

is an odd function, its integral over a symmetric interval

[-a, a]

is zero. Therefore,

\int_{-1}^{1} f(x) dx = 0

. A student who misses this will perform a lengthy partial fraction calculation, but will arrive at the same result.

Question 18

What is the form of the antiderivative of $f(x) = \frac{1}{(x-2)^2(x^2+2x+2)}$ ? (K, L, M, N are constants)

$K\ln|x-2| - \frac{L}{x-2} + M\ln(x^2+2x+2) + N\arctan(x+1) + C$ (correct answer)
$-\frac{K}{x-2} + M\ln(x^2+2x+2) + N\arctan(x) + C$
$K\ln|x-2| + L\ln((x-2)^2) + M\ln(x^2+2x+2) + N\arctan(x+1) + C$
$K\ln|x-2| - \frac{L}{(x-2)^3} + M\ln(x^2+2x+2) + N\arctan(x+1) + C$

Explanation: The decomposition is

\frac{A}{x-2} + \frac{B}{(x-2)^2} + \frac{Cx+D}{x^2+2x+2}

. The quadratic is irreducible as

b^2-4ac = 4-8 = -4 < 0

. Integrating each term:

\int \frac{A}{x-2} dx = A\ln|x-2|

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

. For the quadratic term, we complete the square:

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

. The integral of

\frac{Cx+D}{(x+1)^2+1}

will produce a

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

term and an

\arctan(x+1)

term. Combining these forms gives the structure in answer A.

Question 19

A common mistake when integrating a decomposed fraction like $\int \frac{B}{(x-c)^2} dx$ is to obtain $B\ln((x-c)^2)$ . What is the correct integral and the error's underlying cause?

The correct integral is $B\ln|x-c|$ . The error is failing to apply chain rule correctly to the squared term.
The correct integral is $-\frac{2B}{(x-c)^3}$ . The error is treating the expression as a u-substitution for a logarithm.
The correct integral is $-\frac{B}{x-c}$ . The error is forgetting to use integration by parts for powers greater than one.
The correct integral is $-\frac{B}{x-c}$ . The error is misapplying the logarithm rule, which only works for an exponent of -1. (correct answer)

Explanation: When you encounter integrals involving powers of linear expressions like

(x-c)^n

, the key is recognizing which integration rules apply based on the exponent value. For

\int \frac{B}{(x-c)^2} dx

, rewrite this as

\int B(x-c)^{-2} dx

. Using the power rule for integration, you get

B \cdot \frac{(x-c)^{-1}}{-1} = -\frac{B}{x-c}

. This is the correct answer that option D provides. The common mistake

B\ln((x-c)^2)

occurs because students incorrectly apply the logarithm integration rule. The logarithm rule

\int \frac{1}{u} du = \ln|u|

only works when the denominator has an exponent of exactly -1, not -2. Option D correctly identifies this as misapplying the logarithm rule. Option A is wrong because

B\ln|x-c|

would be the result if the integral were

\int \frac{B}{x-c} dx

, not with the squared denominator. The chain rule explanation doesn't address the core issue. Option B gives an incorrect result (

-\frac{2B}{(x-c)^3}

) that would come from integrating

\frac{B}{x-c}

incorrectly, and u-substitution isn't the relevant concept here. Option C gives the right answer but incorrectly attributes the error to integration by parts, which isn't needed for this type of rational function. Study tip: Remember that

\int \frac{1}{u} du = \ln|u|

only works when the exponent is exactly -1. For any other negative exponent, use the power rule instead.

Question 20

If $\frac{x^3+4}{x^2(x^2+4)} = \frac{A}{x} + \frac{B}{x^2} + \frac{Cx+D}{x^2+4}$ , what is the value of C?

0
1 (correct answer)
-1
4

Explanation: Start with the identity

x^3+4 = Ax(x^2+4) + B(x^2+4) + (Cx+D)x^2

. Expand and group by powers of x:

x^3+4 = (A+C)x^3 + (B+D)x^2 + 4Ax + 4B

. Equating coefficients gives a system of four equations: (1)

A+C = 1

, (2)

B+D=0

, (3)

4A=0

, (4)

4B=4

. From (3),

A=0

. From (4),

B=1

. Substitute

A=0

into (1):

0+C=1 \Rightarrow C=1

. Substitute

B=1

into (2):

1+D=0 \Rightarrow D=-1

. The question asks for C, which is 1.