Partial Fractions: Repeated & Quadratics - Calculus 2

1. 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 $f(x)$?
   Answer: $(x+3)^2(x^2+9)$ 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 $(x+3)^2(x^2+9)$.
2. What is $\int \frac{x}{(x^2 + 1)^2} dx$?
   Answer: $-\frac{1}{2(x^2 + 1)} + C$ 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 $du$.
3. 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?
   Answer: The correct integral is $-\frac{B}{x-c}$. The error is misapplying the logarithm rule, which only works for an exponent of -1. 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.
4. Evaluate the integral $\int \frac{2x+3}{x^2+2x+5} dx$.
   Answer: $\ln(x^2+2x+5) + \frac{1}{2}\arctan\left(\frac{x+1}{2}\right) + C$ 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.
5. 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?
   Answer: 1 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.
6. The partial fraction decomposition of $\frac{4}{x^4-1}$ contains four terms. Which of the following is one of those terms?
   Answer: $\frac{-2}{x^2+1}$ First, factor the denominator completely: $x^4-1 = (x^2-1)(x^2+1) = (x-1)(x+1)(x^2+1)$. The decomposition is $\frac{A}{x-1} + \frac{B}{x+1} + \frac{Cx+D}{x^2+1}$. Solving for the coefficients: $A=1$ (let $x=1$), $B=-1$ (let $x=-1$). This gives $4 = (x+1)(x^2+1) - (x-1)(x^2+1) + (Cx+D)(x^2-1)$. Let $x=0$: $4 = 1 - (-1) + D(-1) \Rightarrow 4 = 2-D \Rightarrow D=-2$. Equating $x^3$ coefficients: $0 = A+B+C = 1-1+C \Rightarrow C=0$. The terms are $\frac{1}{x-1}$, $\frac{-1}{x+1}$, and $\frac{-2}{x^2+1}$. Thus, $\frac{-2}{x^2+1}$ is one of the terms.
7. Evaluate the integral $\int \frac{2x}{(x+1)^2} dx$.
   Answer: $2\ln|x+1| + \frac{2}{x+1} + C$ 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 $(x+1)^{-2}$.
8. If the expression $\frac{4x^2+x+1}{(x-1)(x^2+1)}$ is decomposed into $\frac{A}{x-1} + \frac{Bx+C}{x^2+1}$, what is the value of $A-B$?
   Answer: 2 Start with the equation $4x^2+x+1 = A(x^2+1) + (Bx+C)(x-1)$. Using the cover-up method for $A$, set $x=1$: $4(1)^2+1+1 = A(1^2+1) \Rightarrow 6 = 2A \Rightarrow A=3$. To find C, set $x=0$: $1 = A(1) + C(-1) \Rightarrow 1 = 3 - C \Rightarrow C=2$. To find B, equate the coefficients of $x^2$: $4 = A+B$. Since $A=3$, we have $4 = 3+B \Rightarrow B=1$. The question asks for $A-B$, which is $3-1=2$.
9. Which expression represents the correct partial fraction decomposition of $\frac{1}{(x^2+x+1)^2(x-1)}$?
   Answer: $\frac{Ax+B}{x^2+x+1} + \frac{Cx+D}{(x^2+x+1)^2} + \frac{E}{x-1}$ The denominator has a distinct linear factor $(x-1)$ and a repeated irreducible quadratic factor $(x^2+x+1)^2$. The quadratic is irreducible because its discriminant $b^2-4ac = 1^2 - 4(1)(1) = -3 < 0$. A repeated irreducible quadratic factor $(ax^2+bx+c)^n$ requires n terms of the form $\frac{A_1x+B_1}{ax^2+bx+c} + \frac{A_2x+B_2}{(ax^2+bx+c)^2} + ...$. Thus, the correct form includes terms for $(x^2+x+1)$ and $(x^2+x+1)^2$, both with linear numerators, and a term for $(x-1)$ with a constant numerator.
10. 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?
    Answer: Another natural logarithm with a different argument. 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.
11. What is the correct form of the partial fraction decomposition for the rational expression $\frac{x^3 - 2x + 1}{(x-2)^2 (x^2 + 4)}$?
    Answer: $\frac{A}{x-2} + \frac{B}{(x-2)^2} + \frac{Cx+D}{x^2+4}$ The denominator consists of a repeated linear factor $(x-2)^2$ and an irreducible quadratic factor $(x^2+4)$. A repeated linear factor $(ax+b)^n$ requires n terms in the decomposition: $\frac{A_1}{ax+b} + \frac{A_2}{(ax+b)^2} + ... + \frac{A_n}{(ax+b)^n}$. An irreducible quadratic factor $(ax^2+bx+c)$ requires a linear numerator of the form $Ax+B$. Therefore, the correct decomposition is $\frac{A}{x-2} + \frac{B}{(x-2)^2} + \frac{Cx+D}{x^2+4}$.
12. 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?
    Answer: The term $x^2-4$ is a reducible quadratic and should be factored into $(x-2)(x+2)$. 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 $\frac{Ax+B}{x^2+4} + \frac{C}{x-2} + \frac{D}{x+2}$.
13. 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?
    Answer: Perform polynomial long division, resulting in $\int (x - \frac{1}{x(x^2+1)}) dx$. 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 $x^4+x^2-1$ by $x^3+x$ 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 $\int (x - \frac{1}{x(x^2+1)}) dx$.
14. What is the proper setup for the partial fraction decomposition of the integrand in $\int \frac{x^5+1}{x^2(x^2-2x+5)} dx$?
    Answer: $x+2 + \frac{A}{x} + \frac{B}{x^2} + \frac{Cx+D}{x^2-2x+5}$ The degree of the numerator (5) is greater than the degree of the denominator (4), so polynomial long division must be performed first. The denominator is $x^4-2x^3+5x^2$. Dividing $x^5+1$ by this gives a quotient of $x+2$ and a remainder of $-x^3-10x^2+1$. Thus, the integrand is $x+2 + \frac{-x^3-10x^2+1}{x^2(x^2-2x+5)}$. The remaining rational expression is then decomposed. The denominator has a repeated linear factor $x$ and an irreducible quadratic factor $x^2-2x+5$ (discriminant is $4-20=-16<0$). The proper decomposition form for this fraction is $\frac{A}{x} + \frac{B}{x^2} + \frac{Cx+D}{x^2-2x+5}$. Combining with the polynomial part gives the final setup.
15. Which of the following is the result of $\int \frac{1}{x^2(x^2+1)} dx$?
    Answer: $-\frac{1}{x} - \arctan(x) + C$ The decomposition is $\frac{1}{x^2(x^2+1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{Cx+D}{x^2+1}$. This gives $1 = Ax(x^2+1) + B(x^2+1) + (Cx+D)x^2$. Expanding and equating coefficients of powers of x: $1 = (A+C)x^3 + (B+D)x^2 + Ax + B$. This yields $B=1$, $A=0$, $A+C=0 \Rightarrow C=0$, and $B+D=0 \Rightarrow D=-1$. The integral becomes $\int (\frac{1}{x^2} - \frac{1}{x^2+1}) dx$. Integrating term by term gives $-\frac{1}{x} - \arctan(x) + C$.
16. 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 $\frac{E}{(x^2+1)^2}$?
    Answer: A trigonometric substitution such as $x = \tan(\theta)$. 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.
17. Which of the following integrals would require the use of both repeated linear factors and irreducible quadratic factors in its partial fraction decomposition?
    Answer: $\int \frac{2x - 1}{(x - 3)^2(x^2 + 2x + 5)} dx$ When approaching partial fraction decomposition problems, you need to analyze the denominator to identify what types of factors are present. The question asks specifically for an integral requiring both repeated linear factors and irreducible quadratic factors. Let's examine the denominators systematically. In option D, we have $(x - 3)^2(x^2 + 2x + 5)$. The factor $(x - 3)^2$ is a repeated linear factor (linear factor $x - 3$ appearing twice). For the quadratic $x^2 + 2x + 5$, check if it's irreducible by finding its discriminant: $b^2 - 4ac = 4 - 20 = -16 < 0$. Since the discriminant is negative, this quadratic has no real roots and is irreducible. Therefore, option D contains exactly what we need: both repeated linear factors and irreducible quadratic factors. Now let's eliminate the wrong choices. Option A has $x^2(x^2 + 4)$ - this contains repeated linear factors ($x^2 = (x-0)^2$) and an irreducible quadratic, but $x^2$ isn't typically classified as a "repeated linear factor" in the traditional sense. Option B has $x(x^2 + 6x + 10)^2$ - the quadratic $x^2 + 6x + 10$ is irreducible (discriminant = $36 - 40 = -4 < 0$), but there are no repeated linear factors, only a repeated irreducible quadratic. Option C has $(x^2 + 1)^2$ - this is only a repeated irreducible quadratic with no linear factors. Study tip: Always factor denominators completely and check quadratics' discriminants to distinguish between reducible and irreducible factors. This classification determines your partial fraction setup.
18. 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?
    Answer: $\frac{Ax + B}{x^2 + 1} + \frac{Cx + D}{(x^2 + 1)^2} + \frac{E}{x - 2}$ 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.
19. 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 $P(x)$?
    Answer: Degree 6 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.
20. Evaluate $\int_{-1}^{1} \frac{x^3}{(x^2+2)^2} dx$.
    Answer: 0 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.

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