Linear Partial Fractions - Calculus 2

1. The integral $\int \frac{2x^2 + 7x + 3}{(x + 1)(x + 2)(x + 3)} \, dx$ requires partial fraction decomposition. If the decomposition is $\frac{A}{x + 1} + \frac{B}{x + 2} + \frac{C}{x + 3}$, what is the value of $A + B + C$?
   Answer: $2$ Since the degree of the numerator equals the degree of the denominator minus 1, we can find $A + B + C$ by examining the coefficient of $x^2$ on both sides after clearing denominators, or by substituting a convenient value. Method 1: The coefficient of $x^2$ in the numerator is 2. After clearing denominators: $2x^2 + 7x + 3 = A(x+2)(x+3) + B(x+1)(x+3) + C(x+1)(x+2)$. Expanding the right side, the coefficient of $x^2$ is $A + B + C$. Therefore $A + B + C = 2$. Method 2: We can find each coefficient individually. Setting $x = -1$: $2 - 7 + 3 = A(1)(2) = 2A$, so $A = -1$. Setting $x = -2$: $8 - 14 + 3 = B(-1)(1) = -B$, so $B = 3$. Setting $x = -3$: $18 - 21 + 3 = C(-2)(-1) = 2C$, so $C = 0$. Therefore $A + B + C = -1 + 3 + 0 = 2$.
2. Evaluate $\int \frac{e^{2x}}{e^{2x}-e^x-6} dx$.
   Answer: $\frac{3}{5}\ln|e^x-3| + \frac{2}{5}\ln(e^x+2) + C$ Use the substitution $u = e^x$, so $du = e^x dx$. The integral becomes $\int \frac{e^x \cdot e^x dx}{e^{2x}-e^x-6} = \int \frac{u}{u^2-u-6} du$. The denominator factors as $(u-3)(u+2)$. The partial fraction decomposition is $\frac{u}{(u-3)(u+2)} = \frac{A}{u-3} + \frac{B}{u+2}$. Solving gives $A=3/5$ and $B=2/5$. The integral in $u$ is $\int (\frac{3/5}{u-3} + \frac{2/5}{u+2}) du = \frac{3}{5}\ln|u-3| + \frac{2}{5}\ln|u+2| + C$. Substituting back $u=e^x$, we get $\frac{3}{5}\ln|e^x-3| + \frac{2}{5}\ln(e^x+2) + C$. The absolute value on the second term is not necessary as $e^x+2 > 0$ for all $x$. (B) results from swapping the coefficients A and B. (C) results from an incorrect substitution, such as thinking the numerator is $u^2$ instead of $u$, leading to an improper fraction requiring long division. (D) results from a sign error when solving for B.
3. Which of the following integrals is most suitable for evaluation using the method of linear partial fractions after an appropriate substitution?
   Answer: $\int \frac{1}{\sqrt{x}(x-4)} dx$ (B) After the substitution $u = \sqrt{x}$, we have $x=u^2$ and $dx = 2u du$. The integral transforms into $\int \frac{2u du}{u(u^2-4)} = \int \frac{2}{(u-2)(u+2)} du$, which can be solved using partial fractions with distinct linear factors. (A) The numerator is the derivative of the denominator, so a simple substitution $u=x^2+x+1$ works. Also, the denominator is an irreducible quadratic, so partial fractions with linear factors would not apply. (C) This is an improper rational function. After long division, it becomes $\int (x - \frac{x}{x^2+1}) dx$, which does not require partial fractions. (D) This integral requires integration by parts.
4. To evaluate $\int \frac{x - 7}{(x - 1)(x - 4)} \, dx$, a student sets up the partial fraction decomposition $\frac{x - 7}{(x - 1)(x - 4)} = \frac{A}{x - 1} + \frac{B}{x - 4}$ and correctly finds $A = -2$ and $B = 1$. However, when writing the final answer, the student writes $-2 \ln|x - 1| + \ln|x - 4| + C$. What error did the student make?
   Answer: The student should have $A = 2$ and $B = -1$ instead of the given values Let's check the student's coefficients. Setting up: $\frac{x - 7}{(x - 1)(x - 4)} = \frac{A}{x - 1} + \frac{B}{x - 4}$. Clearing denominators: $x - 7 = A(x - 4) + B(x - 1)$. Setting $x = 1$: $1 - 7 = -6 = A(1 - 4) = -3A$, so $A = 2$. Setting $x = 4$: $4 - 7 = -3 = B(4 - 1) = 3B$, so $B = -1$. The student incorrectly found $A = -2$ and $B = 1$. The correct values are $A = 2$ and $B = -1$, giving the integral $2 \ln|x - 1| - \ln|x - 4| + C$.
5. To evaluate $\int \frac{1}{x^2(x-1)} dx$, a student incorrectly sets up the partial fraction decomposition as $\frac{A}{x} + \frac{B}{x-1}$. If the student proceeds with this incorrect setup, what value of $A+B$ would they find by clearing denominators and comparing the coefficients of the $x$ term?
   Answer: $0$ When you encounter a rational function like $\frac{1}{x^2(x-1)}$, partial fraction decomposition requires accounting for all factors in the denominator. The correct setup should be $\frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-1}$ because the repeated factor $x^2$ needs both $\frac{A}{x}$ and $\frac{B}{x^2}$ terms. However, let's follow the student's incorrect setup: $\frac{1}{x^2(x-1)} = \frac{A}{x} + \frac{B}{x-1}$ Clearing denominators by multiplying both sides by $x^2(x-1)$: $1 = Ax(x-1) + Bx^2$ $1 = Ax^2 - Ax + Bx^2$ $1 = (A+B)x^2 - Ax + 0$ Comparing coefficients: - Constant term: $1 = 0$ (impossible!) - Coefficient of $x$: $0 = -A$, so $A = 0$ - Coefficient of $x^2$: $0 = A + B$, so $0 + B = 0$, giving $B = 0$ Therefore, $A + B = 0 + 0 = 0$, making (C) correct. (A) and (B) represent specific numerical values that might seem plausible but don't result from the coefficient comparison. (D) might seem tempting since the setup is fundamentally flawed (notice the impossible equation $1 = 0$), but the algebra still produces definite coefficient values. Study tip: Always check that your partial fraction setup accounts for repeated factors properly. When you see $x^n$ in the denominator, you need terms for $x^1, x^2, ..., x^n$. If your coefficient comparison leads to contradictions like $1 = 0$, it's a red flag that your initial setup was incorrect.
6. After performing partial fraction decomposition on $\frac{4x + 1}{x^2 - 1}$, the resulting integral becomes $\int \left(\frac{A}{x-1} + \frac{B}{x+1}\right) dx$. Which of the following represents the antiderivative?
   Answer: $\frac{5}{2} \ln|x - 1| + \frac{3}{2} \ln|x + 1| + C$ First, decompose $\frac{4x + 1}{x^2 - 1} = \frac{4x + 1}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}$. Multiplying by $(x-1)(x+1)$: $4x + 1 = A(x+1) + B(x-1)$. Setting $x = 1$: $5 = 2A$, so $A = \frac{5}{2}$. Setting $x = -1$: $-3 = -2B$, so $B = \frac{3}{2}$. Therefore: $\int \frac{4x + 1}{x^2 - 1} dx = \int \left(\frac{5/2}{x-1} + \frac{3/2}{x+1}\right) dx = \frac{5}{2} \ln|x-1| + \frac{3}{2} \ln|x+1| + C$.
7. Let $F(x)$ be the antiderivative of $f(x) = \sec x$. Using the substitution $u=\sin x$ and the method of partial fractions, which of the following is an expression for $F(x)$?
   Answer: $\frac{1}{2}\ln|\frac{1+\sin x}{1-\sin x}| + C$ When you encounter $\int \sec x \, dx$, this is a classic integral that requires a clever approach combining substitution and partial fractions. The key insight is to multiply by a strategic form of 1. To find $F(x) = \int \sec x \, dx$, multiply $\sec x$ by $\frac{1+\sin x}{1+\sin x}$: $\int \sec x \, dx = \int \frac{1}{\cos x} \cdot \frac{1+\sin x}{1+\sin x} \, dx = \int \frac{1+\sin x}{\cos x(1+\sin x)} \, dx = \int \frac{1+\sin x}{1-\sin^2 x} \, dx$ Using the substitution $u = \sin x$, so $du = \cos x \, dx$, this becomes: $\int \frac{1+u}{1-u^2} \, du = \int \frac{1+u}{(1-u)(1+u)} \, du = \int \frac{1}{1-u} \, du$ This integrates to $-\ln|1-u| = \ln|\frac{1}{1-u}| = \ln|\frac{1}{1-\sin x}| = \ln|\frac{1+\sin x}{1-\sin x}|^{1/2} = \frac{1}{2}\ln|\frac{1+\sin x}{1-\sin x}|$ Therefore, answer C is correct. A is wrong because $\ln|1-\sin^2 x| = \ln|\cos^2 x|$ doesn't arise from this method. B has the fraction inverted—it would give the negative of our result. D suggests $\arctan(\sin x)$, which comes from a completely different type of integral. Study tip: For tricky trigonometric integrals like $\int \sec x \, dx$ and $\int \csc x \, dx$, memorize the "multiply by 1" tricks. These standard results appear frequently and the derivations are too time-consuming for exams.
8. The partial fraction decomposition of $\frac{4x^2-3x-4}{x^3-x}$ is given by $\frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1}$. What is the value of the coefficient B?
   Answer: $-3/2$ The denominator factors as $x(x-1)(x+1)$. To find the coefficient B, we can use the Heaviside cover-up method. Multiply the entire rational function by $(x-1)$ and evaluate at $x=1$: $B = \frac{4x^2-3x-4}{x(x+1)} |_{x=1} = \frac{4(1)^2-3(1)-4}{1(1+1)} = \frac{4-3-4}{2} = \frac{-3}{2}$. (A) is the value of A. (B) is the value of C. (D) would be the sum of B and C, which could be mistakenly calculated if a sign error was made.
9. The partial fraction decomposition of $\frac{3x^2 + x + 1}{x(x^2 + 1)}$ requires which of the following forms before integration can proceed?
   Answer: $\frac{A}{x} + \frac{Bx + C}{x^2 + 1}$ When you encounter partial fraction decomposition, you need to break down a rational function into simpler fractions based on the factors in the denominator. The key is matching the correct form to each type of factor. For $\frac{3x^2 + x + 1}{x(x^2 + 1)}$, examine the denominator: it has a linear factor $x$ and an irreducible quadratic factor $x^2 + 1$ (irreducible because it has no real roots). The partial fraction form must account for both factors appropriately. The correct decomposition is $\frac{A}{x} + \frac{Bx + C}{x^2 + 1}$, which is answer choice D. For a linear factor like $x$, you use $\frac{A}{x}$. For an irreducible quadratic factor like $x^2 + 1$, you use $\frac{Bx + C}{x^2 + 1}$ with both a linear term and constant in the numerator. Choice A incorrectly includes $3x$ as a separate polynomial term, which only appears when the degree of the numerator equals or exceeds the degree of the denominator. Choice B treats $x^2 + 1$ as if it were a repeated linear factor, using forms like $\frac{B}{x^2 + 1}$ and $\frac{C}{(x^2 + 1)^2}$, which is wrong for irreducible quadratics. Choice C attempts to factor $x^2 + 1$ over the complex numbers into $(x + i)(x - i)$, but partial fractions work with real factors only. Remember: linear factors get $\frac{A}{\text{factor}}$, irreducible quadratics get $\frac{Bx + C}{\text{quadratic}}$. Always work with the factorization over the real numbers.
10. Evaluate the definite integral $$\int_2^3 \frac{x+5}{x^2+x-2} dx$$
    Answer: $\ln(16/5)$ First, factor the denominator: $x^2+x-2 = (x+2)(x-1)$. The integrand can be decomposed using partial fractions: $\frac{x+5}{(x+2)(x-1)} = \frac{A}{x+2} + \frac{B}{x-1}$. To find the coefficients, we use the equation $x+5 = A(x-1) + B(x+2)$. Setting $x=1$ gives $6 = 3B$, so $B=2$. Setting $x=-2$ gives $3 = -3A$, so $A=-1$. The integral becomes $\int_2^3 (\frac{-1}{x+2} + \frac{2}{x-1}) dx$. Evaluating this gives $[-\ln|x+2| + 2\ln|x-1|]_2^3$. Plugging in the limits: $(-\ln(5) + 2\ln(2)) - (-\ln(4) + 2\ln(1)) = -\ln(5) + 2\ln(2) + \ln(4) = -\ln(5) + 2\ln(2) + 2\ln(2) = 4\ln(2) - \ln(5) = \ln(2^4) - \ln(5) = \ln(16/5)$. (B) results from a sign error during the evaluation of the definite integral. (C) results from swapping the coefficients A and B. (D) is the reciprocal of (C).
11. The partial fraction decomposition of $\frac{3x^2 - 7x + 1}{x^3 - 3x^2 + 2x}$ has the form $\frac{A}{x} + \frac{B}{x-1} + \frac{C}{x-2}$. What is the value of $A+B+C$?
    Answer: $3$ To find $A+B+C$, one can either solve for each coefficient individually or use a shortcut. The shortcut involves recombining the partial fraction form: $\frac{A(x-1)(x-2) + Bx(x-2) + Cx(x-1)}{x(x-1)(x-2)}$. The numerator is $A(x^2-3x+2) + B(x^2-2x) + C(x^2-x)$. The coefficient of the $x^2$ term is $A+B+C$. By equating this to the coefficient of the $x^2$ term in the original numerator, $3x^2 - 7x + 1$, we find that $A+B+C = 3$. Alternatively, solving for each coefficient: $A = \frac{1}{(-1)(-2)} = 1/2$. $B = \frac{3-7+1}{1(-1)} = \frac{-3}{-1} = 3$. $C = \frac{3(4)-7(2)+1}{2(1)} = \frac{12-14+1}{2} = -1/2$. Then $A+B+C = 1/2 + 3 - 1/2 = 3$. (B) is the coefficient of the x-term. (C) would be the result if there was a sign error in calculating B. (D) is the constant term.
12. Evaluate $\int \frac{\cos(x)}{\sin^2(x) - 5\sin(x) + 6} dx$.
    Answer: $\ln|\frac{\sin(x)-3}{\sin(x)-2}| + C$ When you encounter an integral with trigonometric functions where the denominator involves powers of sin(x) or cos(x), look for opportunities to use substitution. Here, the key insight is that the derivative of sin(x) is cos(x), which appears in the numerator. Let $u = \sin(x)$, so $du = \cos(x)dx$. The integral becomes: $\int \frac{du}{u^2 - 5u + 6}$ Factor the denominator: $u^2 - 5u + 6 = (u-2)(u-3)$. This gives us: $\int \frac{du}{(u-2)(u-3)}$ Use partial fraction decomposition: $\frac{1}{(u-2)(u-3)} = \frac{A}{u-2} + \frac{B}{u-3}$ Solving for A and B: $1 = A(u-3) + B(u-2)$. Setting $u=2$ gives $A=-1$, and setting $u=3$ gives $B=1$. So: $\int \frac{du}{(u-2)(u-3)} = \int \left(\frac{-1}{u-2} + \frac{1}{u-3}\right)du = -\ln|u-2| + \ln|u-3| + C$ Using logarithm properties: $\ln|u-3| - \ln|u-2| = \ln\left|\frac{u-3}{u-2}\right|$ Substituting back $u = \sin(x)$: $\ln\left|\frac{\sin(x)-3}{\sin(x)-2}\right| + C$ Option A uses arctangent, which would arise from a quadratic that doesn't factor nicely. Option B has the fraction flipped. Option C lacks the fraction form entirely, missing the logarithm property application. Remember: When you see cos(x) in the numerator with powers of sin(x) in the denominator, try $u = \sin(x)$ substitution first.
13. When decomposing $\frac{2x + 3}{(x + 1)(x - 2)}$ into partial fractions $\frac{A}{x + 1} + \frac{B}{x - 2}$, which system of equations correctly determines the coefficients $A$ and $B$?
    Answer: $A + B = 2$ and $-2A + B = 3$ When you encounter a rational function with distinct linear factors in the denominator, partial fraction decomposition breaks it into simpler fractions that are easier to integrate or manipulate. To find the system of equations for the coefficients, start with the decomposition: $\frac{2x + 3}{(x + 1)(x - 2)} = \frac{A}{x + 1} + \frac{B}{x - 2}$ Multiply both sides by $(x + 1)(x - 2)$ to clear denominators: $2x + 3 = A(x - 2) + B(x + 1)$ Expand the right side: $2x + 3 = Ax - 2A + Bx + B = (A + B)x + (-2A + B)$ For this equation to hold for all values of $x$, the coefficients of like terms must be equal. Comparing coefficients of $x$: $A + B = 2$. Comparing constant terms: $-2A + B = 3$. This gives us the system in choice B: $A + B = 2$ and $-2A + B = 3$. Choice A incorrectly sets the coefficient of $x$ to 0 instead of 2. Choice C has the wrong sign in the second equation—it should be $-2A + B$, not $A - 2B$. Choice D makes both errors: wrong coefficient of $x$ (should be 2, not 0) and wrong form of the constant term equation. Study tip: Always expand $A(x-2) + B(x+1)$ carefully and match coefficients systematically. The coefficient of $x$ on the left must equal the coefficient of $x$ on the right, and the same for constants.
14. The velocity of a particle is given by $v(t) = \frac{10}{25-t^2}$ for $t \in [0, 4]$. If the particle's position at $t=0$ is $s(0) = \ln(5)$, what is its position at $t=4$?
    Answer: $\ln(45)$ Position $s(t)$ is the integral of velocity $v(t)$. We need to calculate $\int \frac{10}{25-t^2} dt$. Using partial fractions: $\frac{10}{(5-t)(5+t)} = \frac{A}{5-t} + \frac{B}{5+t}$. Solving for coefficients: $A = \frac{10}{5+5} = 1$ and $B = \frac{10}{5-(-5)} = 1$. So, $s(t) = \int (\frac{1}{5-t} + \frac{1}{5+t}) dt = -\ln|5-t| + \ln|5+t| + C = \ln|\frac{5+t}{5-t}| + C$. Using the initial condition $s(0)=\ln 5$: $\ln 5 = \ln|\frac{5}{5}| + C = \ln(1) + C = C$. So $C=\ln 5$. The position function is $s(t) = \ln|\frac{5+t}{5-t}| + \ln 5$. At $t=4$, the position is $s(4) = \ln|\frac{5+4}{5-4}| + \ln 5 = \ln(9) + \ln(5) = \ln(45)$. (A) is the value of $s(4)$ if $C=0$. (C) results from a factor error in the integration. (D) results from subtracting $\ln 5$ instead of adding.
15. Consider the integral $\int \frac{x + 2}{x^2 - 3x + 2} \, dx$. After factoring the denominator and setting up partial fractions, what is the coefficient of $\ln|x - 1|$ in the final antiderivative?
    Answer: $-3$ When you encounter a rational function where the degree of the numerator is less than the degree of the denominator, partial fraction decomposition is your go-to technique. This method breaks down complex fractions into simpler ones that are easier to integrate. First, factor the denominator: $x^2 - 3x + 2 = (x-1)(x-2)$. Now set up partial fractions: $\frac{x+2}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2}$ To find the coefficients, multiply both sides by $(x-1)(x-2)$: $x+2 = A(x-2) + B(x-1)$ Using substitution: when $x = 1$, we get $3 = A(-1)$, so $A = -3$. When $x = 2$, we get $4 = B(1)$, so $B = 4$. Therefore: $\frac{x+2}{(x-1)(x-2)} = \frac{-3}{x-1} + \frac{4}{x-2}$ Integrating: $\int \frac{x+2}{x^2-3x+2} dx = \int \frac{-3}{x-1} dx + \int \frac{4}{x-2} dx = -3\ln|x-1| + 4\ln|x-2| + C$ The coefficient of $\ln|x-1|$ is $-3$, making (B) correct. Choice (A) gives $-1$, likely from incorrectly setting up the partial fractions. Choice (C) gives $1$, probably from a sign error in the coefficient calculation. Choice (D) gives $3$, which would result from forgetting the negative sign when $A = -3$. Study tip: Always verify your partial fraction setup by expanding back to check your coefficients before integrating. The substitution method for finding coefficients is usually the most reliable approach.
16. To evaluate $\int \frac{3x + 7}{(x - 1)(x + 2)} \, dx$ using partial fractions, which of the following represents the correct decomposition and subsequent integration?
    Answer: $\frac{10}{3} \ln|x - 1| - \frac{1}{3} \ln|x + 2| + C$ Setting up partial fractions: $\frac{3x + 7}{(x - 1)(x + 2)} = \frac{A}{x - 1} + \frac{B}{x + 2}$. Multiplying both sides by $(x - 1)(x + 2)$: $3x + 7 = A(x + 2) + B(x - 1)$. Setting $x = 1$: $10 = 3A$, so $A = \frac{10}{3}$. Setting $x = -2$: $1 = -3B$, so $B = -\frac{1}{3}$. Therefore: $\int \frac{3x + 7}{(x - 1)(x + 2)} \, dx = \int \frac{10/3}{x - 1} \, dx + \int \frac{-1/3}{x + 2} \, dx = \frac{10}{3} \ln|x - 1| - \frac{1}{3} \ln|x + 2| + C$. Choice B swaps the coefficients, choice C has incorrect signs, and choice D uses the original numerator coefficients without proper decomposition.
17. Which of the following is the antiderivative of $f(x) = \frac{x^3 + 2x^2 - x - 5}{x^2-1}$?
    Answer: $\frac{x^2}{2} + 2x + \frac{3}{2}\ln|\frac{x+1}{x-1}| + C$ The rational function is improper since the degree of the numerator (3) is greater than the degree of the denominator (2). First, perform polynomial long division: $(x^3 + 2x^2 - x - 5) \div (x^2-1)$ gives a quotient of $x+2$ and a remainder of $-3$. So, $f(x) = x+2 - \frac{3}{x^2-1}$. Now, decompose the fractional part: $\frac{3}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}$. Solving for A and B gives $A=3/2$ and $B=-3/2$. The integral is $\int (x+2 - (\frac{3/2}{x-1} - \frac{3/2}{x+1})) dx$ which equals $\frac{x^2}{2} + 2x - \frac{3}{2}\ln|x-1| + \frac{3}{2}\ln|x+1| + C$. Using logarithm properties, this simplifies to $\frac{x^2}{2} + 2x + \frac{3}{2}\ln|\frac{x+1}{x-1}| + C$. (B) results from a sign error in the partial fraction coefficients or the remainder. (C) is the result of forgetting to integrate the quotient $x+2$ from the long division. (D) results from an error in the long division, leading to an incorrect remainder.
18. Consider the partial fraction decomposition $\frac{5x-12}{x(x-4)} = \frac{A}{x} + \frac{B}{x-4}$. What is the value of $A - B$?
    Answer: $1$ Partial fraction decomposition breaks down complex rational functions into simpler fractions that are easier to integrate or manipulate. When you see a fraction with a factored denominator like this, you're setting up unknown constants (A and B) that you'll solve for systematically. To find A and B, multiply both sides by the common denominator $x(x-4)$: $5x - 12 = A(x-4) + Bx$ You can solve this using two efficient methods. The substitution method: substitute $x = 0$ to get $-12 = A(-4)$, so $A = 3$. Then substitute $x = 4$ to get $20 - 12 = B(4)$, so $B = 2$. Alternatively, expand the right side: $5x - 12 = Ax - 4A + Bx = (A+B)x - 4A$. Comparing coefficients: the coefficient of $x$ gives $A + B = 5$, and the constant term gives $-4A = -12$, so $A = 3$. Therefore $B = 2$. With $A = 3$ and $B = 2$, we get $A - B = 3 - 2 = 1$. Choice (A) $-5$ would result from incorrectly calculating $B - A$ instead of $A - B$. Choice (B) $5$ comes from mistakenly finding $A + B$ rather than their difference. Choice (C) $2$ is simply the value of $B$ alone, not the difference. Remember: always double-check your partial fraction decomposition by combining the fractions back together. If $\frac{3}{x} + \frac{2}{x-4}$ gives you back the original expression, you've solved correctly.
19. What is the value of the definite integral $\int_1^5 \frac{dx}{x(x+3)}$?
    Answer: $\frac{1}{3} \ln(\frac{5}{2})$ When you encounter a rational function with a denominator that factors into distinct linear terms, partial fraction decomposition is your key tool. This technique breaks down complex fractions into simpler parts that are easier to integrate. First, decompose $\frac{1}{x(x+3)}$ into partial fractions: $\frac{1}{x(x+3)} = \frac{A}{x} + \frac{B}{x+3}$. To find the constants, multiply both sides by $x(x+3)$: $1 = A(x+3) + Bx$. Setting $x = 0$ gives $A = \frac{1}{3}$, and setting $x = -3$ gives $B = -\frac{1}{3}$. So $\frac{1}{x(x+3)} = \frac{1/3}{x} - \frac{1/3}{x+3}$. Now integrate: $\int_1^5 \frac{dx}{x(x+3)} = \int_1^5 \left(\frac{1/3}{x} - \frac{1/3}{x+3}\right)dx = \frac{1}{3}[\ln|x| - \ln|x+3|]_1^5$ Using logarithm properties, this becomes $\frac{1}{3}[\ln(\frac{x}{x+3})]_1^5 = \frac{1}{3}\left[\ln(\frac{5}{8}) - \ln(\frac{1}{4})\right] = \frac{1}{3}\ln(\frac{5/8}{1/4}) = \frac{1}{3}\ln(\frac{5}{2})$ Answer choice A gives $\frac{1}{3}\ln(10)$, which would result from incorrectly adding the logarithms instead of subtracting. Choice B, $\frac{1}{3}\ln(2)$, comes from calculation errors in the evaluation step. Choice C, $3\ln(\frac{5}{2})$, results from forgetting the $\frac{1}{3}$ factor from the partial fraction decomposition. Remember: partial fractions turn products in denominators into differences of logarithms, and always double-check your arithmetic when evaluating definite integrals at the bounds.
20. In which step did the student make their first error?
    Answer: Step 3 The first error occurs in Step 3. The general antiderivative of $\frac{1}{x-a}$ is $\ln|x-a|$, not $\ln(x-a)$. While the arguments of the logarithms are positive on the interval of integration $[4, 6]$, making the final numerical result correct, the written expression for the antiderivative is mathematically imprecise and incorrect. An antiderivative must be valid over the function's entire domain. For example, using this antiderivative on an interval where $x-3$ is negative would lead to taking the logarithm of a negative number. Step 1 is correct; the factorization and decomposition form are correct. Step 2 is correct; the coefficients are calculated properly. Step 4 correctly evaluates the (incorrect) expression from Step 3.

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