Partial Fractions - IB Mathematics Analysis and Approaches

1. For a non-zero constant $c$, the expression $\frac{c}{x(x+c)}$ is decomposed into $\frac{A}{x} + \frac{B}{x+c}$. Find the value of $A-B$.
   Answer: 2 When you encounter a rational expression that needs to be decomposed into partial fractions, you're working with one of the most systematic algebraic techniques in IB Math AA. The goal is to break down a complex fraction into simpler parts that are easier to work with, especially for integration or solving equations. To decompose $\frac{c}{x(x+c)} = \frac{A}{x} + \frac{B}{x+c}$, multiply both sides by the common denominator $x(x+c)$: $c = A(x+c) + Bx$ Expanding the right side: $c = Ax + Ac + Bx = (A+B)x + Ac$ For this equation to hold for all values of x, the coefficients of like terms must be equal: - Coefficient of x: $A + B = 0$ - Constant term: $Ac = c$ From the second equation, since $c \neq 0$, we get $A = 1$. From the first equation: $1 + B = 0$, so $B = -1$. Therefore, $A - B = 1 - (-1) = 2$. Looking at the wrong answers: (A) 0 would result if you incorrectly concluded $A = B$, missing the sign difference. (B) 1 comes from finding $A$ correctly but forgetting to subtract $B$, or miscalculating the final step. (C) 2c appears if you don't fully solve for the coefficients and leave them in terms of c. Remember this pattern: in partial fraction decomposition, always set up the equation by clearing denominators, then match coefficients systematically. The arithmetic often involves simple integers, even when the original expression contains parameters like c.
2. If $\frac{2x-11}{(x-4)(x+1)} = \frac{A}{x-4} + \frac{B}{x+1}$, find the product $AB$.
   Answer: -\frac{39}{25} We have the identity $2x-11 = A(x+1) + B(x-4)$. To find A, let $x=4$: $2(4)-11 = A(4+1) \implies 8-11 = 5A \implies -3 = 5A \implies A = -\frac{3}{5}$. To find B, let $x=-1$: $2(-1)-11 = B(-1-4) \implies -2-11 = -5B \implies -13 = -5B \implies B = \frac{13}{5}$. The product $AB$ is: $AB = (-\frac{3}{5})(\frac{13}{5}) = -\frac{39}{25}$. Distractor Rationale: A) -39/5: An error in multiplying the denominators (5*1 instead of 5*5). C) 39/25: A sign error in calculating either A or B. D) 39/5: Combination of a sign error and a denominator multiplication error.
3. The function $f(x) = \frac{2x+3}{x^2-x-2}$ is decomposed into $\frac{A}{x-2} + \frac{B}{x+1}$. Find the value of $A^2+B^2$.
   Answer: \frac{50}{9} When you encounter a rational function that needs to be decomposed into partial fractions, you're working with one of the most systematic techniques in algebra. The key is setting up the decomposition correctly based on the factors in the denominator. First, factor the denominator: $x^2-x-2 = (x-2)(x+1)$. This gives us the partial fraction form $\frac{2x+3}{(x-2)(x+1)} = \frac{A}{x-2} + \frac{B}{x+1}$. To find A and B, multiply both sides by $(x-2)(x+1)$: $2x+3 = A(x+1) + B(x-2)$. Now use substitution to solve for each coefficient. When $x = 2$: $2(2)+3 = A(3) + B(0)$, so $7 = 3A$ and $A = \frac{7}{3}$. When $x = -1$: $2(-1)+3 = A(0) + B(-3)$, so $1 = -3B$ and $B = -\frac{1}{3}$. Therefore: $A^2 + B^2 = \left(\frac{7}{3}\right)^2 + \left(-\frac{1}{3}\right)^2 = \frac{49}{9} + \frac{1}{9} = \frac{50}{9}$ Answer A $\frac{50}{9}$ is correct. Answer B $\frac{8}{3}$ likely comes from incorrectly calculating just $A^2$. Answer C (4) might result from using integer approximations instead of exact fractions. Answer D (50) occurs when students forget to square the denominators, calculating $49 + 1 = 50$ instead of $\frac{49}{9} + \frac{1}{9}$. Always work with exact fractions in partial fraction decomposition—decimal approximations will lead you astray on these precision-dependent problems.
4. Find the value of $\int_3^5 \frac{2x}{x^2-4} dx$.
   Answer: \ln($\frac{21}{5}$) This integral can be solved in two ways. Method 1: u-substitution. Let $u = x^2-4$, then $du = 2x\,dx$. The integral becomes $\int \frac{1}{u} du = \ln|u| + C = \ln|x^2-4| + C$. Evaluating the definite integral: $[\ln|x^2-4|]_3^5 = \ln(5^2-4) - \ln(3^2-4) = \ln(21) - \ln(5) = \ln(\frac{21}{5})$. Method 2: Partial Fractions. $\frac{2x}{x^2-4} = \frac{2x}{(x-2)(x+2)} = \frac{A}{x-2} + \frac{B}{x+2}$ $2x = A(x+2) + B(x-2)$ Let $x=2$: $4 = 4A \implies A=1$. Let $x=-2$: $-4 = -4B \implies B=1$. The integral is $\int_3^5 (\frac{1}{x-2} + \frac{1}{x+2}) dx = [\ln|x-2| + \ln|x+2|]_3^5$. $= [\ln|(x-2)(x+2)|]_3^5 = [\ln|x^2-4|]_3^5$, which leads to the same result $\ln(\frac{21}{5})$. Distractor Rationale: A) $\ln(\frac{15}{7})$: This result is obtained if one incorrectly finds $B=-1$, leading to an integral of $\ln|\frac{x-2}{x+2}|$. B) $\ln(4)$: Result of an incorrect log calculation like $\ln(21-5)=\ln(16)=4\ln(2)$ or other arithmetic error. D) $\ln(\frac{5}{3})$: This would be the result of integrating $\frac{2}{x}$, not $\frac{2x}{x^2-4}$.
5. Find the exact value of the integral $\int_2^4 \frac{6}{x^2-1} dx$.
   Answer: 3\ln($\frac{9}{5}$) First, decompose the integrand using partial fractions. $\frac{6}{x^2-1} = \frac{6}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}$ $6 = A(x+1) + B(x-1)$ Let $x=1$: $6 = 2A \implies A=3$. Let $x=-1$: $6 = -2B \implies B=-3$. So the integral becomes: $\int_2^4 (\frac{3}{x-1} - \frac{3}{x+1}) dx = [3\ln|x-1| - 3\ln|x+1|]_2^4$ Using logarithm properties, this is $[3\ln|\frac{x-1}{x+1}|]_2^4$. Evaluate at the limits: $3\ln(\frac{4-1}{4+1}) - 3\ln(\frac{2-1}{2+1}) = 3\ln(\frac{3}{5}) - 3\ln(\frac{1}{3})$ $= 3(\ln(\frac{3}{5}) - \ln(\frac{1}{3})) = 3\ln(\frac{3/5}{1/3}) = 3\ln(\frac{3}{5} \times 3) = 3\ln(\frac{9}{5})$. Distractor Rationale: A) 3ln(5/9): Sign error, likely from reversing the subtraction of logs. C) 3ln(5): Common error if B is incorrectly found to be 3 instead of -3. The integral would become $\int \frac{3}{x-1} + \frac{3}{x+1} dx = 3\ln|x^2-1|$, which evaluates to $3(\ln 15 - \ln 3) = 3\ln 5$. D) 6ln(5): A combination of the previous error and another mistake.
6. The expression $\frac{px+q}{x^2-4}$ is equivalent to $\frac{A}{x-2} + \frac{B}{x+2}$. Find an expression for $A$ in terms of $p$ and $q$.
   Answer: \frac{2p+q}{4} We start with the identity $px+q = A(x+2) + B(x-2)$. To find $A$, we want to eliminate the term with $B$. We can do this by setting $x-2=0$, i.e., $x=2$. Substitute $x=2$ into the identity: $p(2)+q = A(2+2) + B(2-2)$ $2p+q = A(4) + B(0)$ $2p+q = 4A$ $A = \frac{2p+q}{4}$. Distractor Rationale: A) 2p+q: Forgetting to divide by 4. B) $\frac{2p-q}{4}$: This is the expression for B. D) $\frac{q-2p}{4}$: A sign error in the expression for B.
7. The expression $\frac{8}{(2x+1)(2x-3)}$ is written in the form $\frac{A}{2x+1} + \frac{B}{2x-3}$. Find the value of $B-A$.
   Answer: 4 We start with the identity $8 = A(2x-3) + B(2x+1)$. To find A, let $2x+1=0$, which means $x = -1/2$. $8 = A(2(-1/2)-3) + B(0) \implies 8 = A(-1-3) \implies 8 = -4A \implies A=-2$. To find B, let $2x-3=0$, which means $x = 3/2$. $8 = A(0) + B(2(3/2)+1) \implies 8 = B(3+1) \implies 8 = 4B \implies B=2$. The question asks for $B-A$. $B-A = 2 - (-2) = 4$. Distractor Rationale: A) -4: This is $A-B$. B) 0: This is $A+B$. This could also result from a sign error where a student calculates $A=2$ or $B=-2$. C) 2: This is just the value of B.
8. The expression $\frac{kx-1}{(x-3)(x+2)}$ is written as $\frac{A}{x-3} + \frac{B}{x+2}$. Given that $A=2$, find the value of $k$.
   Answer: \frac{11}{3} We start with the identity $kx-1 = A(x+2) + B(x-3)$. We can find an expression for A using the cover-up method. Let $x=3$: $k(3)-1 = A(3+2) + B(0) \implies 3k-1 = 5A$. So, $A = \frac{3k-1}{5}$. We are given that $A=2$. $2 = \frac{3k-1}{5}$ $10 = 3k-1$ $11 = 3k$ $k = \frac{11}{3}$. Alternatively, we can use simultaneous equations. We are given $A=2$. Equating coefficients of $x$: $k = A+B = 2+B$. Equating constant terms: $-1 = 2A-3B = 2(2)-3B = 4-3B$. From the constant terms: $-5 = -3B \implies B=5/3$. Then, substitute B back into the equation for k: $k = 2 + 5/3 = 6/3 + 5/3 = 11/3$. Distractor Rationale: A) 1: Results from an error in the cover-up denominator, e.g., using $3-2=1$ instead of $3-(-2)=5$. B) 3: Results from an algebraic slip, e.g., $3k-1=10 \implies 3k=9 \implies k=3$. D) 9/2: Results from confusing A and B, setting the expression for B equal to 2.
9. When $\frac{13x-12}{6x^2-x-2}$ is expressed as partial fractions, the result is $\frac{A}{3x-2} + \frac{B}{2x+1}$. Find $A-B$.
   Answer: -\frac{47}{7} First, confirm the denominator factorization: $(3x-2)(2x+1) = 6x^2+3x-4x-2 = 6x^2-x-2$. This is correct. We have the identity $13x-12 = A(2x+1) + B(3x-2)$. To find A, let $3x-2=0$, so $x=2/3$. $13(\frac{2}{3}) - 12 = A(2(\frac{2}{3})+1) \implies \frac{26}{3} - \frac{36}{3} = A(\frac{4}{3}+\frac{3}{3}) \implies -\frac{10}{3} = A(\frac{7}{3}) \implies A = -\frac{10}{7}$. To find B, let $2x+1=0$, so $x=-1/2$. $13(-\frac{1}{2}) - 12 = B(3(-\frac{1}{2})-2) \implies -\frac{13}{2} - \frac{24}{2} = B(-\frac{3}{2}-\frac{4}{2}) \implies -\frac{37}{2} = B(-\frac{7}{2}) \implies B = \frac{37}{7}$. We need to find $A-B$: $A-B = -\frac{10}{7} - \frac{37}{7} = -\frac{47}{7}$. Distractor Rationale: B) -27/7: Results from a sign error, e.g., $A=10/7$, giving $10/7 - 37/7 = -27/7$. C) 27/7: This is $A+B$ with a sign error on A, or $B-A$ with a sign error on B. Also the value of $B+A$ is $-10/7 + 37/7 = 27/7$. D) 47/7: This is $B-A$.
10. Let $f(x) = \frac{5}{(x+2)(x-3)}$. Find the exact value of $\int_4^5 f(x) dx$.
    Answer: \ln($\frac{12}{7}$) When you encounter a rational function like this in an integral, the key technique is partial fraction decomposition. The function $f(x) = \frac{5}{(x+2)(x-3)}$ has distinct linear factors in the denominator, so you can split it into simpler fractions. Set up the decomposition: $\frac{5}{(x+2)(x-3)} = \frac{A}{x+2} + \frac{B}{x-3}$ Multiplying both sides by $(x+2)(x-3)$: $5 = A(x-3) + B(x+2)$ Solving for the constants: when $x = 3$, we get $5 = 5B$, so $B = 1$. When $x = -2$, we get $5 = -5A$, so $A = -1$. Therefore: $f(x) = \frac{-1}{x+2} + \frac{1}{x-3}$ Now integrate: $\int_4^5 f(x)dx = \int_4^5 \left(\frac{-1}{x+2} + \frac{1}{x-3}\right)dx = [-\ln|x+2| + \ln|x-3|]_4^5$ Evaluating: $[-\ln(7) + \ln(2)] - [-\ln(6) + \ln(1)] = -\ln(7) + \ln(2) + \ln(6) = \ln\left(\frac{2 \cdot 6}{7}\right) = \ln\left(\frac{12}{7}\right)$ This confirms answer A is correct. Answer B gives the reciprocal, suggesting a sign error in the partial fractions. Answer C represents a calculation error, likely from incorrect partial fraction coefficients. Answer D results from overlooking one of the logarithmic terms in the evaluation. Remember: partial fraction decomposition is your go-to method for integrating rational functions with factorizable denominators. Always double-check your coefficients before integrating.
11. Let $f(x) = \frac{1}{x^2-3x+2}$. The partial fraction decomposition of $f(x)$ is $\frac{A}{x-2} + \frac{B}{x-1}$. Which statement correctly describes the behavior of the graph of $y=f(x)$ near its vertical asymptotes?
    Answer: The function approaches +$\infty$ as $x \to 1^{-}$ and +$\infty$ as $x \to 2^{+}$. First, find the partial fraction decomposition. The denominator is $x^2-3x+2 = (x-1)(x-2)$. The vertical asymptotes are at $x=1$ and $x=2$. $\frac{1}{(x-2)(x-1)} = \frac{A}{x-2} + \frac{B}{x-1}$ $1 = A(x-1) + B(x-2)$ Let $x=2$: $1 = A(2-1) \implies A=1$. Let $x=1$: $1 = B(1-2) \implies B=-1$. So, $f(x) = \frac{1}{x-2} - \frac{1}{x-1}$. The behavior near the asymptotes is determined by the signs of the coefficients A and B. Near $x=1$, the term $-\frac{1}{x-1}$ dominates. As $x \to 1^{-}$ (e.g., x=0.99), $x-1$ is a small negative number. So $-\frac{1}{x-1}$ is a large positive number. Thus, $f(x) \to +\infty$. As $x \to 1^{+}$ (e.g., x=1.01), $x-1$ is a small positive number. So $-\frac{1}{x-1}$ is a large negative number. Thus, $f(x) \to -\infty$. Near $x=2$, the term $\frac{1}{x-2}$ dominates. As $x \to 2^{-}$ (e.g., x=1.99), $x-2$ is a small negative number. So $\frac{1}{x-2}$ is a large negative number. Thus, $f(x) \to -\infty$. As $x \to 2^{+}$ (e.g., x=2.01), $x-2$ is a small positive number. So $\frac{1}{x-2}$ is a large positive number. Thus, $f(x) \to +\infty$. Checking the options: B) The function approaches +$\infty$ as $x \to 1^{-}$ (correct) and +$\infty$ as $x \to 2^{+}$ (correct). This statement is true.
12. Given $\frac{k}{(x-a)(x+a)} = \frac{P}{x-a} + \frac{Q}{x+a}$, where $k$ and $a$ are non-zero constants. Which of the following statements is always true?
    Answer: P + Q = 0 From the equation, we have the identity $k = P(x+a) + Q(x-a)$. To find P, let $x=a$: $k = P(a+a) + Q(a-a) \implies k = P(2a) \implies P = \frac{k}{2a}$. To find Q, let $x=-a$: $k = P(-a+a) + Q(-a-a) \implies k = Q(-2a) \implies Q = -\frac{k}{2a}$. Now we check the given statements: A) $P=Q$ is false, as $\frac{k}{2a} = -\frac{k}{2a}$ implies $k=0$ or $a=0$, but we are given they are non-zero. B) $P+Q = \frac{k}{2a} + (-\frac{k}{2a}) = 0$. This is always true. C) $PQ = (\frac{k}{2a})(-\frac{k}{2a}) = -\frac{k^2}{4a^2}$. This is not generally 1. D) $P-Q = \frac{k}{2a} - (-\frac{k}{2a}) = \frac{2k}{2a} = \frac{k}{a}$. This is not a general relationship like $P+Q=0$, but an expression for $P-Q$. The question asks what is always true as a relationship between P and Q. $P+Q=0$ is a fundamental relationship derived from the structure of the problem.
13. Given $\int_1^k \frac{1}{x(x+1)} dx = \ln(3/2)$, find the value of $k$, where $k>1$.
    Answer: 3 First, decompose the integrand using partial fractions. $\frac{1}{x(x+1)} = \frac{A}{x} + \frac{B}{x+1}$ $1 = A(x+1) + Bx$ Let $x=0$: $1 = A(1) \implies A=1$. Let $x=-1$: $1 = B(-1) \implies B=-1$. The integral becomes $\int_1^k (\frac{1}{x} - \frac{1}{x+1}) dx$. This integrates to $[\ln|x| - \ln|x+1|]_1^k = [\ln|\frac{x}{x+1}|]_1^k$. Now, evaluate at the limits: $\ln(\frac{k}{k+1}) - \ln(\frac{1}{1+1}) = \ln(\frac{k}{k+1}) - \ln(\frac{1}{2})$. Using log rules, this is $\ln(\frac{k/(k+1)}{1/2}) = \ln(\frac{2k}{k+1})$. We are given that the result is $\ln(3/2)$. $\ln(\frac{2k}{k+1}) = \ln(\frac{3}{2})$ $\frac{2k}{k+1} = \frac{3}{2}$ $4k = 3(k+1)$ $4k = 3k+3$ $k=3$. Distractor Rationale: A) 2: A simple guess or calculation error. C) 4: A simple guess or calculation error. D) 5: A simple guess or calculation error.
14. Let $f(x) = \frac{4x}{(1-x)(1+3x)}$. In the series expansion for $f(x)$ in ascending powers of $x$, what is the coefficient of the $x$ term?
    Answer: 4 First, decompose $f(x)$ into partial fractions. $\frac{4x}{(1-x)(1+3x)} = \frac{A}{1-x} + \frac{B}{1+3x}$ $4x = A(1+3x) + B(1-x)$ Let $x=1$: $4(1) = A(1+3) \implies 4=4A \implies A=1$. Let $x=-1/3$: $4(-1/3) = B(1 - (-1/3)) \implies -4/3 = B(4/3) \implies B=-1$. So $f(x) = \frac{1}{1-x} - \frac{1}{1+3x} = (1-x)^{-1} - (1+3x)^{-1}$. Now, find the first few terms of the series expansion for each part: $(1-x)^{-1} = 1+x+x^2+...$ $(1+3x)^{-1} = 1-3x+(3x)^2-... = 1-3x+9x^2-...$ Substitute back: $f(x) = (1+x+...) - (1-3x+...)$ The $x$ term is $x - (-3x) = 4x$. The coefficient is 4. Note: For a function of the form $\frac{kx}{(1-ax)(1-bx)}$, the coefficient of $x$ in the Maclaurin series is always $k$. In this case $k=4$. Distractor Rationale: A) -2: Results from an error like $1+(-3)=-2$, incorrectly combining coefficients. B) 2: Results from a sign error, for instance if $B=1$ was found, giving $1-3=-2$, and then another sign error. D) 3: Not an obvious error path, but a plausible incorrect integer.
15. The integral $\int \frac{5x+5}{x^2+x-6} dx$ can be written in the form $\ln|f(x)|+C$. Find $f(x)$.
    Answer: $(x+3)^2$$(x-2)^3$ First, decompose the integrand into partial fractions. The denominator is $x^2+x-6 = (x+3)(x-2)$. $\frac{5x+5}{(x+3)(x-2)} = \frac{A}{x+3} + \frac{B}{x-2}$ $5x+5 = A(x-2) + B(x+3)$ Let $x=2$: $5(2)+5 = B(2+3) \implies 15 = 5B \implies B=3$. Let $x=-3$: $5(-3)+5 = A(-3-2) \implies -10 = -5A \implies A=2$. The integral is $\int (\frac{2}{x+3} + \frac{3}{x-2}) dx$. This integrates to $2\ln|x+3| + 3\ln|x-2| + C$. Using logarithm properties: $\ln|(x+3)^2| + \ln|(x-2)^3| + C = \ln|(x+3)^2(x-2)^3| + C$. So, $f(x) = (x+3)^2(x-2)^3$. Distractor Rationale: A) $(x+3)^3$$(x-2)^2$: This occurs if the values of A and B are swapped. C) $\frac{(x+3)^2}{(x-2)^3}$: This occurs if B is incorrectly found to be -3. D) $\frac{(x-2)^3}{(x+3)^2}$: This occurs if A is incorrectly found to be -2.
16. A function is given by $f(x) = \frac{4}{x+3} - \frac{1}{x-1}$. Find the equivalent expression for $f(x)$ as a single rational function.
    Answer: \frac{3x-7}{x^2+2x-3} To combine the fractions, we find a common denominator, which is $(x+3)(x-1)$. $f(x) = \frac{4(x-1)}{(x+3)(x-1)} - \frac{1(x+3)}{(x+3)(x-1)}$ $f(x) = \frac{4(x-1) - (x+3)}{(x+3)(x-1)}$ Now, simplify the numerator: $4x - 4 - x - 3 = 3x - 7$ Expand the denominator: $(x+3)(x-1) = x^2 - x + 3x - 3 = x^2+2x-3$ So, $f(x) = \frac{3x-7}{x^2+2x-3}$. Distractor Rationale: A) $\frac{3x-1}{x^2+2x-3}$: A sign error in the numerator: $4x-4-x+3=3x-1$. C) $\frac{5x-1}{x^2+2x-3}$: A sign error in the original expression, combining $\frac{4}{x+3} + \frac{1}{x-1}$, which gives a numerator of $4(x-1)+(x+3)=5x-1$. D) $\frac{3x-7}{x^2-3}$: A common error when multiplying binomials, $(x+3)(x-1) = x^2 - 3$.
17. Let $f(x) = \frac{3-x}{(1-2x)(1+x)}$. When $f(x)$ is expanded as a series in ascending powers of $x$, find the coefficient of $x^2$.
    Answer: 8 First, decompose $f(x)$ into partial fractions: $\frac{3-x}{(1-2x)(1+x)} = \frac{A}{1-2x} + \frac{B}{1+x}$ $3-x = A(1+x) + B(1-2x)$ Let $x=1/2$: $3-1/2 = A(1+1/2) \implies 5/2 = A(3/2) \implies A=5/3$. Let $x=-1$: $3-(-1) = B(1-2(-1)) \implies 4 = B(3) \implies B=4/3$. So, $f(x) = \frac{5}{3}(1-2x)^{-1} + \frac{4}{3}(1+x)^{-1}$. Now, use the binomial expansion $(1+y)^n = 1+ny+\frac{n(n-1)}{2}y^2+...$. $(1-2x)^{-1} = 1 + (-1)(-2x) + \frac{(-1)(-2)}{2}(-2x)^2 + ... = 1+2x+4x^2+...$ $(1+x)^{-1} = 1 + (-1)(x) + \frac{(-1)(-2)}{2}(x)^2 + ... = 1-x+x^2+...$ Substitute these back into the expression for $f(x)$: $f(x) = \frac{5}{3}(1+2x+4x^2+...) + \frac{4}{3}(1-x+x^2+...)$ The coefficient of $x^2$ is $\frac{5}{3}(4) + \frac{4}{3}(1) = \frac{20}{3} + \frac{4}{3} = \frac{24}{3} = 8$. Distractor Rationale: A) 5: This is the sum of the $x^2$ coefficients (4+1) without multiplying by A and B. B) 7: This results from swapping A and B: $\frac{4}{3}(4) + \frac{5}{3}(1) = \frac{16+5}{3} = 7$. D) 16/3: This results from a sign error in the expansion of $(1+x)^{-1}$, for example, getting a coefficient of -1 for $x^2$. Then $\frac{5}{3}(4) + \frac{4}{3}(-1) = \frac{20-4}{3} = \frac{16}{3}$.
18. Given that $\frac{2x+3}{2x^2-x-1} = \frac{A}{2x+1} + \frac{B}{x-1}$, find the value of $B$.
    Answer: \frac{5}{3} First, factor the denominator: $2x^2-x-1 = (2x+1)(x-1)$. So we have the identity $2x+3 = A(x-1) + B(2x+1)$. To find B, we choose a value of $x$ that makes the term with A equal to zero. Let $x=1$. $2(1)+3 = A(1-1) + B(2(1)+1) \implies 5 = A(0) + B(3) \implies 5 = 3B \implies B = \frac{5}{3}$. For completeness, to find A, let $x=-1/2$. $2(-1/2)+3 = A(-1/2-1) + B(0) \implies -1+3 = A(-3/2) \implies 2 = -\frac{3}{2}A \implies A = -\frac{4}{3}$. Distractor Rationale: A) -4/3: This is the value of A. B) -1/3: A possible arithmetic error. D) 5: Forgetting to divide by 3 when solving for B.
19. Using the fact that $\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$, find the sum of the infinite series $\sum_{n=1}^{\infty} \frac{2}{n(n+1)}$.
    Answer: 2 When you encounter an infinite series with a given partial fraction decomposition, you're dealing with a telescoping series—one of the most elegant techniques in calculus for finding exact sums. Start by factoring out the constant from the given series: $\sum_{n=1}^{\infty} \frac{2}{n(n+1)} = 2\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$ Now apply the given identity $\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$ to get: $2\sum_{n=1}^{\infty} \left(\frac{1}{n} - \frac{1}{n+1}\right)$ Write out the first few terms to see the telescoping pattern: $2\left[\left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \cdots\right]$ Notice how consecutive terms cancel: the $-\frac{1}{2}$ from the first term cancels with $+\frac{1}{2}$ from the second term, and so on. The partial sum up to $n$ terms is $2\left(1 - \frac{1}{n+1}\right)$. As $n \to \infty$, this approaches $2(1 - 0) = 2$. Answer choice B (1) would result if you forgot to multiply by the factor of 2. Answer choice C (0) might occur if you incorrectly thought all terms cancel completely. Answer choice D (diverges) would be wrong because telescoping series with vanishing end terms always converge to finite values. The correct answer is A. Study tip: Whenever you see a partial fraction identity provided in a series problem, immediately look for telescoping. Write out several terms explicitly—the cancellation pattern will become obvious and guide you to the sum.

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