Exercise 7.5 · Q4: $\int\frac{x}{(x-1)(x-2)(x-3)}\,dx$

Basics you need

Solution

1. The fraction is proper, with three different linear factors

   The numerator $x$ has degree 1 and the denominator has degree 3, so the fraction is proper. The factors $x-1$⁠, $x-2$⁠, $x-3$ are linear and different.

   These are the two conditions for row 3 of Table 7.2: proper, and three different linear factors.
   Section 7.5. Lesson: partial fractions

2. Write the form of the decomposition

   $$\begin{aligned} &\frac{x}{(x-1)(x-2)(x-3)} \\ &\quad = \frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{x-3} \qquad (1) \end{aligned}$$
   One constant over each different linear factor.
   Table 7.2, row 3. Basics: splitting fractions

3. Clear the denominators

   Multiply (1) by $(x-1)(x-2)(x-3)$⁠:

   $$\begin{aligned} x &= A(x-2)(x-3) \\ &\quad + B(x-1)(x-3) \\ &\quad + C(x-1)(x-2). \qquad (2) \end{aligned}$$
   (1) holds for all $x \ne 1, 2, 3$⁠, so the polynomials in (2) agree at infinitely many points and are equal for every real $x$⁠.
   Remark after Example 11, Section 7.5.

4. Find $A$⁠, $B$⁠, $C$ by putting in the roots

   $x = 1$⁠: $1 = A(1-2)(1-3) = 2A$⁠, so $A = \frac12$⁠.

   $x = 2$⁠: $2 = B(2-1)(2-3) = -B$⁠, so $B = -2$⁠.

   $x = 3$⁠: $3 = C(3-1)(3-2) = 2C$⁠, so $C = \frac32$⁠.

   At each root two of the three terms of (2) contain a zero factor and vanish, leaving one equation in one unknown.

5. Confirm by comparing coefficients

   With $(x-2)(x-3) = x^2 - 5x + 6$⁠, $(x-1)(x-3) = x^2 - 4x + 3$⁠, $(x-1)(x-2) = x^2 - 3x + 2$⁠, the right side of (2) is

   $$(A+B+C)x^2 - (5A+4B+3C)x + (6A+3B+2C).$$

   Compare with $0x^2 + 1x + 0$⁠, using $A = \frac12$⁠, $B = -2$⁠, $C = \frac32$⁠:

   $$\begin{aligned} x^2:&\quad \tfrac12 - 2 + \tfrac32 = 0, \\ x:&\quad -\big(\tfrac52 - 8 + \tfrac92\big) = -(-1) = 1, \\ \text{const}:&\quad 3 - 6 + 3 = 0. \end{aligned}$$
   All three coefficients agree with the numerator $x$⁠, so the constants are right.
   Equal polynomials have equal coefficients. Basics: splitting fractions

6. Integrate the three simple fractions

   $$\begin{aligned} \int f(x)\,dx &= \frac12\int\frac{dx}{x-1} - 2\int\frac{dx}{x-2} \\ &\quad + \frac32\int\frac{dx}{x-3} \\ &= \tfrac12\log|x-1| - 2\log|x-2| \\ &\quad + \tfrac32\log|x-3| + C. \end{aligned}$$
   Integral of a sum, constants outside (Properties (III), (IV), Section 7.2.1); each piece is $\int\frac{dx}{x+k} = \log|x+k|$⁠. The final $C$ is the arbitrary constant of integration, not the $C = \frac32$ of the decomposition.
   Standard formula (xii), Section 7.2. Basics: logarithms and 1/x

7. Check: differentiate the answer

   The derivative is $\frac{1/2}{x-1} - \frac{2}{x-2} + \frac{3/2}{x-3}$⁠. Over the common denominator $(x-1)(x-2)(x-3)$ its numerator is

   $$\begin{aligned} &\tfrac12(x^2-5x+6) - 2(x^2-4x+3) \\ &\quad + \tfrac32(x^2-3x+2) \\ &= 0x^2 + \big(-\tfrac52 + 8 - \tfrac92\big)x + (3 - 6 + 3) \\ &= x. \ \checkmark \end{aligned}$$
   The derivative is $\frac{x}{(x-1)(x-2)(x-3)}$⁠, the integrand. The answer is valid on every interval avoiding $1$⁠, $2$⁠, $3$⁠.
   $\frac{d}{dx}\log|x| = \frac1x$⁠, Section 7.2, list item (xii). Basics: integration undoes differentiation

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