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Chapter 7 · Integrals

Exercise 7.5 · Q22 (MCQ): ∫x dx(x−1)(x−2)\int\frac{x\,dx}{(x - 1)(x - 2)}

Exercise 7.5Integration by partial fractions · Section 7.5
The question

Choose the correct answer.

∫x dx(x−1)(x−2)  equals\int\frac{x\,dx}{(x - 1)(x - 2)} \ \text{ equals}

(A) log⁡∣(x−1)2x−2∣+C\log\left|\frac{(x - 1)^2}{x - 2}\right| + C    (B) log⁡∣(x−2)2x−1∣+C\log\left|\frac{(x - 2)^2}{x - 1}\right| + C
(C) log⁡∣(x−1x−2)2∣+C\log\left|\Big(\frac{x - 1}{x - 2}\Big)^2\right| + C    (D) log⁡∣(x−1)(x−2)∣+C\log|(x - 1)(x - 2)| + C

Exam answer

Let II be the given integral, and let

x(x−1)(x−2)=Ax−1+Bx−2\frac{x}{(x - 1)(x - 2)} = \frac{A}{x - 1} + \frac{B}{x - 2}

Then x=A(x−2)+B(x−1)x = A(x - 2) + B(x - 1)⁠. Putting x=1x = 1⁠: A=−1A = -1⁠. Putting x=2x = 2⁠: B=2B = 2⁠.

I=−log⁡∣x−1∣+2log⁡∣x−2∣+C=log⁡∣(x−2)2x−1∣+C\begin{aligned} I &= -\log|x - 1| \\ &\quad + 2\log|x - 2| + C \\ &= \log\left|\frac{(x - 2)^2}{x - 1}\right| + C \end{aligned}

Answer: (B) log⁡∣(x−2)2x−1∣+C\log\left|\dfrac{(x - 2)^2}{x - 1}\right| + C

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

Split the integrand by Table 7.2 (row 1): x(x−1)(x−2)=−1x−1+2x−2\frac{x}{(x - 1)(x - 2)} = \frac{-1}{x - 1} + \frac{2}{x - 2}⁠. The coefficients −1-1 and 22 become the powers inside the logarithm: −log⁡∣x−1∣+2log⁡∣x−2∣-\log|x - 1| + 2\log|x - 2| =log⁡∣(x−2)2x−1∣= \log\left|\frac{(x - 2)^2}{x - 1}\right|⁠, which is option (B).

Given, to find, and the plan
Given
The integrand x(x−1)(x−2)\frac{x}{(x - 1)(x - 2)}⁠, defined for x≠1,2x \ne 1, 2⁠, and four options.
To find
Which option is ∫x dx(x−1)(x−2)\int\frac{x\,dx}{(x - 1)(x - 2)}⁠.
Plan
(1) Split into partial fractions. (2) Integrate and combine the logarithms. (3) Differentiate back. (4) Test every option by differentiating it.

Basics you need

1Table 7.2, row 1

px+q(x−a)(x−b)=Ax−a+Bx−b\frac{px + q}{(x - a)(x - b)} = \frac{A}{x - a} + \frac{B}{x - b}⁠, a≠ba \ne b⁠. Lesson: partial fractions

2Logarithm laws

klog⁡∣p∣=log⁡∣p∣kk\log|p| = \log|p|^k⁠, log⁡∣p∣+log⁡∣q∣=log⁡∣pq∣\log|p| + \log|q| = \log|pq|⁠, log⁡∣p∣−log⁡∣q∣=log⁡∣pq∣\log|p| - \log|q| = \log\left|\frac pq\right|⁠. Basics: logarithms and 1/x

3Testing an option

An option is correct exactly when its derivative equals the integrand. Basics: integration undoes differentiation

Solution

  1. Split into partial fractions

    The fraction is proper (degree 1 over degree 2) with two different linear factors, so

    x(x−1)(x−2)=Ax−1+Bx−2,x=A(x−2)+B(x−1).\begin{gathered} \frac{x}{(x - 1)(x - 2)} = \frac{A}{x - 1} + \frac{B}{x - 2}, \\ x = A(x - 2) + B(x - 1). \end{gathered}

    Put x=1x = 1⁠: 1=−A1 = -A⁠, so A=−1A = -1⁠. Put x=2x = 2⁠: 2=B2 = B⁠. Coefficient check: A+B=1A + B = 1 ✓ (coefficient of xx⁠) and −2A−B=0-2A - B = 0 ✓ (constant term).

    Row 1 of Table 7.2. The cleared identity holds for every xx⁠, so substituting the roots is allowed and isolates each constant; the coefficient comparison confirms both.
    Table 7.2, row 1. Basics: splitting fractions

  2. Integrate and combine

    ∫(−1x−1+2x−2)dx=−log⁡∣x−1∣+2log⁡∣x−2∣+C=log⁡∣x−2∣2−log⁡∣x−1∣+C=log⁡∣(x−2)2x−1∣+C.\begin{aligned} \int\Big(\frac{-1}{x - 1} + \frac{2}{x - 2}\Big)dx &= -\log|x - 1| + 2\log|x - 2| + C \\ &= \log|x - 2|^2 - \log|x - 1| + C \\ &= \log\left|\frac{(x - 2)^2}{x - 1}\right| + C. \end{aligned}
    Each simple fraction gives a logarithm (standard formula (xii)). The coefficient 2 becomes a square inside the log, and a subtracted log goes to the denominator. This is option (B).
    Standard formula (xii), Section 7.2. Basics: logarithms and 1/x

  3. Check: differentiate the answer

    ddx(2log⁡∣x−2∣−log⁡∣x−1∣)=2x−2−1x−1=2(x−1)−(x−2)(x−1)(x−2)=x(x−1)(x−2). ✓\begin{aligned} &\frac{d}{dx}\big(2\log|x - 2| - \log|x - 1|\big) \\ &= \frac{2}{x - 2} - \frac{1}{x - 1} \\ &= \frac{2(x - 1) - (x - 2)}{(x - 1)(x - 2)} = \frac{x}{(x - 1)(x - 2)}. \ \checkmark \end{aligned}
    Over the common denominator, 2x−2−x+2=x2x - 2 - x + 2 = x⁠, the original numerator.
    Property (I), Section 7.2.1. Basics: integration undoes differentiation

  4. Examine every option

    Write each option as a sum of logarithms and differentiate. At x=4x = 4 the integrand is 43⋅2=23≈0.6667\frac{4}{3\cdot 2} = \frac23 \approx 0.6667⁠.

    • (A) 2log⁡∣x−1∣−log⁡∣x−2∣2\log|x - 1| - \log|x - 2| has derivative 2x−1−1x−2=x−3(x−1)(x−2)\frac{2}{x - 1} - \frac{1}{x - 2} = \frac{x - 3}{(x - 1)(x - 2)}⁠. At x=4x = 4⁠: 16≈0.1667\frac16 \approx 0.1667⁠. The coefficients of the two logs are swapped. Wrong.
    • (B) 2log⁡∣x−2∣−log⁡∣x−1∣2\log|x - 2| - \log|x - 1|⁠: its derivative is the integrand (Step 3). Correct.
    • (C) 2log⁡∣x−1∣−2log⁡∣x−2∣2\log|x - 1| - 2\log|x - 2| has derivative 2x−1−2x−2=−2(x−1)(x−2)\frac{2}{x - 1} - \frac{2}{x - 2} = \frac{-2}{(x - 1)(x - 2)}⁠. At x=4x = 4⁠: −13-\frac13⁠. Wrong.
    • (D) log⁡∣x−1∣+log⁡∣x−2∣\log|x - 1| + \log|x - 2| has derivative 1x−1+1x−2=2x−3(x−1)(x−2)\frac{1}{x - 1} + \frac{1}{x - 2} = \frac{2x - 3}{(x - 1)(x - 2)}⁠. At x=4x = 4⁠: 56≈0.8333\frac56 \approx 0.8333⁠. It treats the integrand as if both coefficients were 1. Wrong.
    An option is an antiderivative exactly when its derivative equals the integrand for every xx in the domain. For (A), (C), (D) the numerators x−3x - 3⁠, −2-2⁠, 2x−32x - 3 are not identically xx⁠, and one sample point where they differ (here x=4x = 4⁠) is enough to reject each.
Result ∫x dx(x−1)(x−2)=log⁡∣(x−2)2x−1∣+C,\int\frac{x\,dx}{(x - 1)(x - 2)} = \log\left|\frac{(x - 2)^2}{x - 1}\right| + C,

for x≠1,2x \ne 1, 2⁠. The correct option is (B). (Textbook answer: B.)

Check it with numbers

Decomposition at x=4x = 4⁠: 43⋅2=23\frac{4}{3\cdot 2} = \frac23 and −13+22=23-\frac13 + \frac22 = \frac23 (≈0.6667\approx 0.6667⁠). ✓

Definite integral from 3 to 4 (continuous there): F(4)−F(3)=log⁡43−log⁡12F(4) - F(3) = \log\frac43 - \log\frac12 =log⁡83≈0.9808293= \log\frac83 \approx 0.9808293⁠. A numerical integration over [3,4][3, 4] gives 0.98082930.9808293 too. ✓

Common mistakes
  1. Choosing (A). Mixing up which factor gets which constant: B=2B = 2 belongs to x−2x - 2⁠, so x−2x - 2 is the squared factor.
  2. Choosing (D). log⁡∣(x−1)(x−2)∣\log|(x - 1)(x - 2)| would need numerator 2x−32x - 3⁠, the derivative of the denominator; here the numerator is xx⁠.
  3. Sign of AA⁠. At x=1x = 1⁠: 1=A(1−2)1 = A(1 - 2)⁠, so A=−1A = -1⁠, not 11⁠.
What to take away
  1. Partial-fraction coefficients become powers inside a combined logarithm.
  2. For an MCQ on integrals, differentiating each option is a complete test.
  3. Substituting the roots of the factors is the fastest way to find the constants.

Every fact used, and where it comes from

FactUsed inSource
Ax−a+Bx−b\frac{A}{x - a} + \frac{B}{x - b}Step 1Table 7.2, row 1, Section 7.5
∫dxx−a=log⁡∣x−a∣+C\int\frac{dx}{x - a} = \log|x - a| + CStep 2Standard formula (xii), Section 7.2
Laws of logarithmsSteps 2, 4Basics: logarithms and 1/x
F′=fF' = f tests an antiderivativeSteps 3, 4Property (I), Section 7.2.1
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