Integrating Using Linear Partial Fractions: Practice Questions (Calculus BC 6.12)

Answer: (A). x² + x − 2 = (x − 1)(x + 2). Write 2x + 7 = A(x + 2) + B(x − 1). At x = 1: 9 = 3A, so A = 3. At x = −2: 3 = −3B, so B = −1. Check: 3(x + 2) − (x − 1) = 2x + 7. ✓

• (B) puts the two constants over the wrong factors. Recombined, the top is 2x − 5.
• (C) loses the minus sign from −3B = 3. Recombined, the top is 4x + 5.
• (D) substitutes the roots into the top only and forgets to divide by the value of the other factor (9 and 3 instead of 9/3 and 3/(−3)).

Answer: (A). x² + 2x − 3 = (x − 1)(x + 3). From 4 = A(x + 3) + B(x − 1): x = 1 gives A = 1, and x = −3 gives B = −1. So the integral is ln|x − 1| − ln|x + 3| + C = ln|(x − 1)/(x + 3)| + C.

• (B) treats the integrand as if it were the derivative of the bottom over the bottom. Its derivative is 8(x + 1)/(x² + 2x − 3).
• (C) has A and B the wrong way round. Its derivative is −4/((x − 1)(x + 3)), the negative of the integrand.
• (D) loses the minus sign on B. Its derivative is (2x + 2)/((x − 1)(x + 3)).

Answer: (A). 2/((x − 1)(x + 1)) = 1/(x − 1) − 1/(x + 1). On [2, 4] both factors are positive, so the antiderivative is ln((x − 1)/(x + 1)). Then [ln((x − 1)/(x + 1))] from 2 to 4 = ln(3/5) − ln(1/3) = ln(9/5) ≈ 0.588.

• (B) evaluates at the upper limit only. A negative value for a positive integrand is a warning sign.
• (C) uses A = 2 and B = −2: it puts the top, 2, over each factor without dividing by the value of the other factor (2 at x = 1, and −2 at x = −1). This doubles the answer.
• (D) swaps the signs of A and B, giving the negative of the correct value.

Answer: (A). Write 7 = A(x − 3) + B(2x + 1). At x = 3: 7 = 7B, so B = 1. At x = −½: 7 = A(−7/2), so A = −2. The integrand is −2/(2x + 1) + 1/(x − 3). Since ∫ 1/(2x + 1) dx = ½ ln|2x + 1|, the integral is −ln|2x + 1| + ln|x − 3| + C = ln|(x − 3)/(2x + 1)| + C.

• (B) finds A and B correctly but forgets the chain-rule factor ½ when integrating −2/(2x + 1).
• (C) has the signs reversed; its derivative is the negative of the integrand.
• (D) takes the log of the expanded denominator. Its derivative is 7(4x − 5)/(2x² − 5x − 3), which is not the integrand.

1. Factor. x³ − x = x(x − 1)(x + 1): three different linear factors. The top has degree 2 and the bottom degree 3, so the fraction is proper.
2. Set up. (4x² + x − 1)/(x(x − 1)(x + 1)) = A/x + B/(x − 1) + C/(x + 1), so 4x² + x − 1 = A(x − 1)(x + 1) + Bx(x + 1) + Cx(x − 1).
3. Substitute the roots.
   • x = 0: −1 = A(−1)(1), so A = 1.
   • x = 1: 4 = B(1)(2), so B = 2.
   • x = −1: 2 = C(−1)(−2), so C = 1.
4. Integrate. ∫ (4x² + x − 1)/(x³ − x) dx = ln|x| + 2 ln|x − 1| + ln|x + 1| + C.

Check. 1/x + 2/(x − 1) + 1/(x + 1) has top (x − 1)(x + 1) + 2x(x + 1) + x(x − 1) = (x² − 1) + (2x² + 2x) + (x² − x) = 4x² + x − 1. ✓ Also, A + B + C = 4 matches the x² coefficient.

Suggested mark points (3): 1 for the correct factorisation and decomposition form with three constants; 1 for A = 1, B = 2, C = 1; 1 for the correct antiderivative with absolute values and + C.

Common error: using x(x² − 1) and trying A/x + B/(x² − 1). The factor x² − 1 is not linear; factor it fully first.

(a) The top and bottom both have degree 2, so f is not a proper fraction. A sum A/(x − 3) + B/(x + 2) can never produce the x² term on top, so you must divide first.

(b) x² + 2x − 10 = 1 · (x² − x − 6) + (3x − 4), because (x² − x − 6) + (3x − 4) = x² + 2x − 10. So f(x) = 1 + (3x − 4)/(x² − x − 6).

(c) x² − x − 6 = (x − 3)(x + 2). From 3x − 4 = A(x + 2) + B(x − 3): x = 3 gives 5 = 5A, so A = 1; x = −2 gives −10 = −5B, so B = 2. (3x − 4)/(x² − x − 6) = 1/(x − 3) + 2/(x + 2).

(d) On [4, 5], x − 3 and x + 2 are positive, so there is no asymptote in the interval. ∫ from 4 to 5 of f(x) dx = [x + ln(x − 3) + 2 ln(x + 2)] from 4 to 5 = (5 + ln 2 + 2 ln 7) − (4 + ln 1 + 2 ln 6) = 1 + ln 2 + 2 ln(7/6), which is 1 + ln(49/18), about 2.001.

Total: 5 points. Acceptable alternatives: matching coefficients in (c) (A + B = 3, 2A − 3B = −4); any correct equivalent exact form in (d), such as 1 + ln(49/18). A decimal alone does not earn the last point, because this part is no-calculator.

(a) 900 = A(t + 10) + B(t + 1). At t = −1: 900 = 9A, so A = 100. At t = −10: 900 = −9B, so B = −100. R(t) = 100/(t + 1) − 100/(t + 10).

(b) For t ≥ 0 both factors are positive. ∫ from 0 to 8 of R(t) dt = [100 ln(t + 1) − 100 ln(t + 10)] from 0 to 8 = 100[ln(9/18) − ln(1/10)] = 100 ln(10/2) = 100 ln 5 thousand users ≈ 160.944 thousand.

(c) Yes. Users at the end of week 8 = 20 + ∫ from 0 to 8 of R(t) dt = 20 + 100 ln 5 ≈ 180.944 thousand, which is more than 175 thousand. The integral of the rate gives the users gained, so the starting 20 thousand must be added before comparing.

Total: 5 points. Acceptable alternatives: a correct answer to (c) carried forward from an incorrect (b); a calculator’s numerical integral for the decimal in (b), but the exact value still needs the decomposition. Common error in (c): comparing 160.944 with 175 and answering “no”, forgetting the 20 thousand starting users.